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Maths M.C.Q (1 Marks)

Principle of Mathematical Induction · CBSE STD 11 Science (CBSE - English Medium). 185 questions.

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M.C.Q (1 Marks)

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185 questions · auto-graded multiple-choice test.

MCQ 11 Mark
If $x^{2 n-1}+y^{2 n-1}$ is divisible by $x + y,$ if $n$ is:
  • ✓
    a positive integer
  • B
    an even positive integer
  • C
    an odd positive integer
  • D
    None of these
Answer
Correct option: A.
a positive integer
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MCQ 21 Mark
The nth terms of the series $3 + 7 + 13 + 21 +……….$ is
  • A
    $4n – 1$
  • B
    $2n + 1$
  • ✓
    $n^2 + n + 1$
  • D
    $n + 2$
Answer
Correct option: C.
$n^2 + n + 1$
Concept:
Let $S = 3 + 7 + 13 + 21 +……….a_{n-1}+ a_n...(1)$
and $S = 3 + 7 + 13 + 21 +……….a_{n-1}+ a_n...(2)$
Substracting Equation $(1)$ and equation $(2)$ we get,
$S – S = 3 + (7 + 13 + 21 +……….a_{n-1}+ a_n) – (3 + 7 + 13 + 21 +……….a_{n-1}+ a_n)$
$\Rightarrow 0 = 3 + (7 – 3) + (13 – 7) + (21 – 13) + ……….+ (a_n– a_{n-1}) – a_n$
$\Rightarrow 0 = 3 + \{4 + 6 + 8 + ……(n-1)\ce{terms}\} – a_n$
$\Rightarrow a_n= 3 + \{4 + 6 + 8 + ……(n-1)\ce{terms}\}$
$4 + 6 + 8 + ……(n-1)$ terms is an Arithmatic Progression with first term $= 4,$ common difference $= 2$ and no. of terms $= n -1$
$\text{a}_\text{n}=3+\frac{\text{n-1}}{2}\times(2\times4+(\text{n}-1-1)\times2)$
$\text{a}_\text{n}=3+\frac{\text{n-1}}{2}\times(8+(\text{n}-2)\times2)$
$\Rightarrow\text{a}_\text{n}=3(\text{n}-1)\times(4+\text{n}-2)$
$\Rightarrow\text{a}_\text{n}=3(\text{n}-1)\times(\text{n}-2)$
$\Rightarrow\text{a}_\text{n}=\text{n}^2+\text{n}+1$
So, the $n_{th}$ term is $n^2+ n + 1$
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MCQ 31 Mark
Let $P(n)$ be a statement and $P(n)=P(n+1)\forall n \in N,$ then $P(n)$ is true for what values of $n?$
  • ✓
    For all $n$
  • B
    For all $n>1$
  • C
    For all $n>m , m$ being a fixed positive integer
  • D
    Nothing can be said
Answer
Correct option: A.
For all $n$
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MCQ 41 Mark
If $P(n) = 2 + 4 + ......+ 2n, n\ \epsilon\ N,$ then $P(k) = k(k + 1) + 2 \Rightarrow P(k) = k(k + 1) + 2$ for all $k\  \epsilon\  N. S$ we can conclude that $P(n) = n(n + 1) + 2$ for
  • A
    all $n\ \epsilon\ N$
  • B
    $n > 1$
  • C
    $n > 2$
  • ✓
    nothing can be said
Answer
Correct option: D.
nothing can be said
Concepts:
Suppose there is a given statement $P(n)$ involving the natural number n such that
The statement is true for $n = 1,$
i.e., $P(1)$ is true, and
If the statement is true for $n = k ($where $k$ is some positive integer$),$ then the statement is also true for $n = k + 1,$
i.e., the truth of $P(k)$ implies the truth of $P(k + 1)$.
Then, $P(n)$ is true for all natural numbers $n$
Calculation:
Given
$P(n) = 2 + 4 + ......+ 2n$
Put $n = 1$
$P(1) = 2$
Hence, $P(n) = n(n + 1) + 2$ is not true for $n = 1$
So, The Principle of Mathematical Induction is not applicable and nothing can be said about the validity of the statement $P(n) = n(n + 1) + 2$
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MCQ 51 Mark
For natural number $n, 2^n, (n – 1) ! < ,$ if:
  • A
    $n < 2$
  • ✓
    $n > 2$
  • C
    $n ≥ 2$
  • D
    never
Answer
Correct option: B.
$n > 2$
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MCQ 61 Mark
Let $P(n) = 2^{3n}− 7n − 1$ then $P(n)$ is divisible by:
  • A
    $63$
  • B
    $36$
  • ✓
    $49$
  • D
    $25$
Answer
Correct option: C.
$49$
$P(n) = 2^{3n}− 7n − 1 = −1 − 7n + (1 + 7)^n$
$\Rightarrow \ce{P(n)=-1-7 n + \left(1 + n C_1 7 + n C_2 7^2 + ... + n C_n 7^n\right)=n C_2 7^2 + ... + n C_n 7^n}$
$\Rightarrow \ce{P(n)=7^2\left({nC}_2 + {nC}_3 7 + ... + {nC}_{n} 7^{n}-2\right)}$
Therefore, $P(n)$ is divisible by $49.$
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MCQ 71 Mark
For all positive integers $n$, the number $n(n^2 − 1)$ is divisible by:
  • A
    $36$
  • B
    $24$
  • ✓
    $6$
  • D
    $16$
Answer
Correct option: C.
$6$
Given,
number $= n(n^2 - 1)$
Let $n = 1, 2, 3, 4….$
$n(n^2 - 1) = 1(1 - 1) = 0$
$n(n^2 - 1) = 2(4 - 1) = 2 \times 3 = 6$
$n(n^2 - 1) = 3(9 - 1) = 3 \times 8 = 24$
$n(n^2 - 1) = 4(16 - 1) = 4 \times 15 = 60$
Since all these numbers are divisible by $6$ for $n = 1, 2, 3,..........$
So, the given number is divisible $6$
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MCQ 81 Mark
Let $P(m)$ be the statement $m^2 > 100$, the statement $P(k + 1)$ will be true if:
  • A
    $P(1)$ is true
  • B
    $P(2)$ is true
  • ✓
    $P(k)$ is true
  • D
    none of these
Answer
Correct option: C.
$P(k)$ is true
$P(r)$ is true
$ \Rightarrow r^2>100$
$\Rightarrow r^2+2 r+1>100+2 r+1 $
$ \Rightarrow(r+1)^2>100 $
$ \Rightarrow P(r+1) $ is true as
$r^2+(2 r+1)>r^2>100$
$\Rightarrow P(k+1)$ is true $($say $r=k)$
$P(k + 1)$ is true when every $p(k)$ is
So, In order to prove that $P(k + 1)$ is true.
It is sufficient to consider $P(k)$ is true.
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MCQ 101 Mark
If $10^{3n}+ 2^{4k+1}× 9 + k,$ is divisible by $11,$ then what is the least positive value of $k?$
  • A
    $7$
  • B
    $6$
  • C
    $8$
  • ✓
    $10$
Answer
Correct option: D.
$10$
$P(n) = 10^{3n} + 2^{4k+1}\times 9 + k$
$P(1) = 10^3+ 2^5\times 9 + k$
$P(1) = 1000 + 288 + k$
$P(1) = 1288 + k$
When $1288$ is divided by $11$, the remainder is $1.$
Therefore, $1287$ is divisible by $11.$
The next number that is divisible is $1298.$
$k = 1298 - 1288$
$k = 10$
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MCQ 111 Mark
If $p(n): 2n < (1 \times 2 \times 3 \times ... \times n).$ Then the smallest positive integer for which $p(n)$ is true is:
  • A
    $1$
  • B
    $2$
  • C
    $3$
  • ✓
    $4$
Answer
Correct option: D.
$4$
The smallest positive integer for which $P(n)$ is $4.$
$P(4) = 2^4 < (1 \times 2 \times 3 \times ... \times 4)$
$P(4) = 16 < 24.$
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MCQ 131 Mark
By principle of mathematical induction, $2^{4n-1}$  is divisible by which of the following?
  • ✓
    $8$
  • B
    $3$
  • C
    $5$
  • D
    $7$
Answer
Correct option: A.
$8$
$P(n)=2^{4 n-1} P(1)=2^3=8$
Let us assume $P(k)$ is divisible by $8$ and can be written as $8c,$
where $c$ is any integer.
$ P(k)=2^{4 k-1}=8 c $
$ P(k+1)=2^{4(k+1)-1} $
$ P(k+1)=2^{4 k+3} $
$ P(k+1)=2^4 \times 2^{4 k-1} $
$ P(k+1)=2^4 \times 8 c $
Clearly, $P(k + 1)$ is divisible by $2, 4, 8$ and $16.$
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MCQ 141 Mark
Let $P(n)$ be statement $2n < n!$. Where n is a natural number, then $P(n)$ is true for:
  • A
    all $n$
  • B
    all $n > 2$
  • ✓
    all $n > 3$
  • D
    none of these
Answer
Correct option: C.
all $n > 3$
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MCQ 151 Mark
For each $n \ N \in , 10^{2n-1}+ 1$ is divisible by
  • ✓
    $11$
  • B
    $13$
  • C
    $9$
  • D
    None of these
Answer
Correct option: A.
$11$
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MCQ 161 Mark
If $S_n$ is divisible for every $n,$ then $S_n$ is:
  • ✓
    $> 0$
  • B
    $> 1$
  • C
    $> 5$
  • D
    None of these
Answer
Correct option: A.
$> 0$
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MCQ 171 Mark
Let $S(k) = 1 + 3 + 5 + (2k – 1) = 3 + K^2$. Then, which of the following is true?
  • A
    S(1) is correct
  • ✓
    S(k) ÞS(k + 2)
  • C
    S(k) ÞS(k + 1)
  • D
    Principle of mathematical induction can be used to prove the formula
Answer
Correct option: B.
S(k) ÞS(k + 2)
b.  S(k) ÞS(k + 1)
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MCQ 181 Mark
The inequality $n \ !>2^{n-1}$ is true for
  • ✓
    $n > 2$
  • B
    $n\in N$
  • C
    $n > 3$
  • D
    None of these
Answer
Correct option: A.
$n > 2$
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MCQ 191 Mark
For $n \in N, \big(\frac{1}{5}\big)\text{n}^5+\big(\frac{1}{3}\big)\text{n}^3+\big(\frac{7}{15}\big)$ is:
  • A
    an integer
  • ✓
    a natural number
  • C
    a positive fraction
  • D
    None of these
Answer
Correct option: B.
a natural number
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MCQ 201 Mark
$n^3+ 5n$ is divisible by which of the following?
  • ✓
    $3$
  • B
    $5$
  • C
    $7$
  • D
    $11$
Answer
Correct option: A.
$3$
$P(n) = n^3 + 5n$
$P(1) = 1 + 5$
$P(1) = 6$
We assume the $P(k)$ is true and divisible by $6.$
$P(k) = k^3+ 5k4$ is divisible by $6$ and can be written as $6c$ or $3 \times 2c$
We need to prove that $P(k + 1)$ is divisible by $6$
$ P(k + 1) = (k + 1)^3+ 5(k + 1)4$
$ P(k + 1) = k^3+ 1 + 3k^2+ 3k + 5k + 5$
$ P(k + 1) = (k^3+ 5k) + 3k^2+ 3k + 6$
$ P(k + 1) = 6c + 3(k^2+ k + 2)$
 $ P(k + 1) = (3 \times 2c) + 3(k^2+ k + 2)$
Therefore, $P(k + 1)$ is definitely divisible by $3$
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MCQ 211 Mark
$n(n + 1) (n + 5)$ is a multiple of $........$ for all $n \in N$
  • A
    $2$
  • ✓
    $3$
  • C
    $5$
  • D
    $7$
Answer
Correct option: B.
$3$
Let $P(n): n(n + 1)(n + 5)$ is a multiple of $3.$
For $n = 1,$ the given expression becomes $(1 \times 2 \times 6) = 12,$
which is a multiple of $3.$
So, the given statement is true for $n = 1,$
i.e. $P(1)$ is true.
Let $P(k)$ be true. Then,
$P(k): k(k + 1)(k + 5)$ is a multiple of $3$
$\Rightarrow K(k + 1) (k + 5) = 3 m$ for some natural number $m, …… (i)$
Now, $(k + 1) (k + 2) (k + 6) = (k + 1) (k + 2)k + 6(k + 1) (k + 2)$
$= k(k + 1) (k + 2) + 6(k + 1) (k + 2)$
$= k(k + 1) (k + 5 – 3) + 6(k + 1) (k + 2)$
$= k(k + 1) (k + 5) – 3k(k + 1) + 6(k + 1) (k + 2)$
$= k(k + 1) (k + 5) + 3(k + 1) (k + 4) [$on simplification$]$
$= 3m + 3(k + 1 ) (k + 4) [$using $(i)]$
$= 3[m + (k + 1) (k + 4)],$ which is a multiple of $3$
$\Rightarrow P(k + 1) (k + 1 ) (k + 2) (k + 6)$ is a multiple of $3$
$\Rightarrow P(k + 1)$ is true, whenever $P(k)$ is true.
Thus, $P(1)$ is true and $P(k + 1)$ is true, whenever $P(k)$ is true.
Hence, by the principle of mathematical induction, $P(n)$ is true for all $n \in N$
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MCQ 221 Mark
If $\forall\text{m}\in\text{N},$ then $11^{m+2}+ 12^{2m-1}$ is divisible by:
  • A
    $121$
  • B
    $132$
  • ✓
    $133$
  • D
    None of these
Answer
Correct option: C.
$133$
To find the divisor of $11^{m+2}+ 12^{2m-1}$ by mathematic induction, the first step is to check for the smallest natural number,
i.e; for $m = 1.$
So, this reduces to $11^3+ 12^1$ or $11^4+ 1$.
So, the number when divided by $11$ leaves remainder $1.$
So, we can knock out options $A$ and $B$ as $121$ as well as $132$ are both divisible by $11$ and
hence their multiples will always be divisible by $11.$
Now, we have to check the divisibility of $11^{m+2}+ 12^{2m-1}$ by $133.$
For $m = 1, 11^4+ 1$ is not divisible by $133.$
So, we can knock out option $C.$
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MCQ 231 Mark
Let $P (n)$ be the statement $3n > nn.$ If $P (n)$ is true, then $P (n + 1)$ is
  • ✓
    True
  • B
    False
  • C
    Not determined
  • D
    None of the above
Answer
Correct option: A.
True
Concepts:
Suppose there is a given statement $P(n)$ involving the natural number $n$ such that
The statement is true for $n = 1,$
i.e., $p(1)$ is true, and
If the statement is true for $n = k ($where $k$ is some positive integer$),$
then the statement is also true for $n = k + 1,$
i.e., the truth of $P(k)$ implies the truth of $P(k + 1).$
Then, $P(n)$ is true for all natural numbers $n$
Calculation:
Given: $P (n)$ is true
For $n = 1$
$P (1): 3^1 > 1^1$
$\Rightarrow 3 > 1$
$\therefore$ It is true for $n = 1$
It is given that $P (k)$ is true,
so $P (k)$ is true for $n = k$
$P(K) \Rightarrow 3^k > k^k$
Now,
$P(k +1) = 3^{k+1} > (k + 1)^{k+1}$
$3^k.3^1 > (k + 1)^k (k + 1)$
By the principle for mathematics production,
$P(n + 1)$ is true,
​​​​​​​when $P(n)$ is true
$\therefore P(n + 1)$ is true.
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MCQ 241 Mark
For any natural number $n, 7^n – 2^n$ is divisible by
  • A
    $3$
  • B
    $4$
  • ✓
    $5$
  • D
    $7$
Answer
Correct option: C.
$5$
Given, $ 7^n – 2^n $
Let $n = 1$ 
$7^n – 2^n$
$= 7^1 – 2^1$
$= 7 – 2$
$= 5 $
which is divisible by $5$
Let $n = 2$
$ 7^n – 2^n$
$= 72 – 22$
$= 49 – 4$
$= 45$
which is divisible by $5$
Let $n = 3$
$ 7^n – 2^n$
$= 7^3 – 2^3$
$= 343 – 8$
$= 335$
which is divisible by $5$
Hence, for any natural number $n, 7^n – 2^n $ is divisible by $5$
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MCQ 251 Mark
For every natural number $k,$ which of the following is true?
  • ✓
    $\ce{(mn)k = mknk}$
  • B
    $\ce{mk2 = n + 1}$
  • C
    $\ce{(m + n)k = k + 1}$
  • D
    $\ce{mkn = mnk}$
Answer
Correct option: A.
$\ce{(mn)k = mknk}$
Concepts:
Suppose there is a given statement $P(n)$ involving the natural number $n$ such that
The statement is true for $n = 1,$
i.e., $P(1)$ is true, and
If the statement is true for $n = k ($where $k$ is some positive integer$),$ then the statement is also true for $n = k + 1,$
i.e., the truth of $P(k)$ implies the truth of $P(k + 1).$
Then, $P(n)$ is true for all natural numbers $n$​​​​​​​
Calculation:
Given:
For, $k = 1$
Option $1,$
$\ce{(mn)^1= m^1n^1, mn = mn (LHS = RHS)}$
hence it is true
Let us assume that the statement is true for, $k = p$
$\ce{(mn)^p= m^pn^p}$
Multiplying the above equation with $mn,$ we get mn we get,
$\ce{(mn)p + 1 = mp +1 np + 1}$
$\ce{mp + 1 np + 1 = mp + 1 np + 1}$
Hence the given expression is true for every natural number $k.$
Option $2, 3$ and $4$ does not satisfy for $k = 1,$
Hence Option $1$ is correct.
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MCQ 291 Mark
What is the sum of $12 + 22 + 32 + ... +\ n2\ ?$
  • ✓
    $\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{2}$
  • B
    $\frac{\text{n}(\text{n+1)}}{6}$
  • C
    $\frac{\text{n}(\text{n+1}+2\text{n+1)}}{6}$
  • D
    $\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{3}$
Answer
Correct option: A.
$\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{2}$
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MCQ 301 Mark
If $49^n+ 16^n+ k$ is divisible by $64$ for $n \in N,$ then the least negative integral value of $k$ is:
  • ✓
    $- 1$
  • B
    $- 2$
  • C
    $- 3$
  • D
    $- 4$
Answer
Correct option: A.
$- 1$
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MCQ 311 Mark
For all $n\ \epsilon \ N, (1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9}).......(1+(\frac{2\text{n+1)}}{2}))$ is equal to:
  • A
    $\frac{(\text{n+1)}^2}{2}$
  • B
    $\frac{(\text{n+1)}^3}{3}$
  • ✓
    $(\text{n+1)}^2$
  • D
    $\text{none}\text{ of}\text{ these}$
Answer
Correct option: C.
$(\text{n+1)}^2$
Concepts:
Suppose there is a given statement $P(n)$ involving the natural number $n$ such that
The statement is true for $n = 1,$
i.e., $p(1)$ is true, and
If the statement is true for $n = k ($where $k$ is some positive integer$)$, then the statement is also true for $n = k + 1,$
i.e., the truth of $P(k)$ implies the truth of $P(k + 1).$
Then, $P(n)$ is true for all natural numbers $n$​​​​​​​
Calculation:
Given:
Let $P(n)$ be defined as
$\text{p}\text{(n)}=\Big(1+\frac{3}{1}\Big)\Big(1+\frac{5}{4}\Big)\Big(1+\frac{7}{9}\Big).......\Big(1+\Big(\frac{2\text{n+1)}}{\text{n}^2}\Big)\Big)=(\text{k+1})^2$
Put $n = 1$
$\text{p}(1)=\Big(1+\frac{3}{1}\Big)=(1+1)^2$
$4 = 4 P(1)$ is true
Let it is true for $n = k$
$\text{p}\text{(n)}=\Big(1+\frac{3}{1}\Big)\Big(1+\frac{5}{4}\Big)\Big(1+\frac{7}{9}\Big).......\Big(1+\Big(\frac{2\text{n+1)}}{\text{n}^2}\Big)\Big)=(\text{k+1})^2 \ \dots(1)$
for $n = k + 1$
$\text{p}\text{(n)}=\Big(1+\frac{3}{1}\Big)\Big(1+\frac{5}{4}\Big)\Big(1+\frac{7}{9}\Big).......\Big(1+\Big(\frac{2\text{n+1)}}{\text{n}^2}\Big)\Big)$
$=(\text{k+1})^2]\Big(1+\frac{2\text{k+1+2}}{(\text{k+1)}^2}\Big)$
$=(\text{k+1})^2]\Big(1+\frac{2\text{k+1+3}}{(\text{k+1)}^2}\Big)$
Using Equation $(1)$
$\text{(k+1)}^2\Big[\frac{(\text{k+1)}^2+2\text{k+3}}{(\text{k+1)}}\Big]$
$=\text{k}^2+2\text{k}+1+2\text{k}+3$
$\text{(k+2)}=[(\text{k}+1)+1]^2$
Therefore, $P(k +1)$ is true,
Hence From the Principle of Mathematical Induction,
the statement is true for all natural numbers $n.$
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MCQ 321 Mark
For a positive integer $n,$ let a $\text{(n)}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2^\text{n-1}}$ Then:
  • ✓
    $a(100) ≤ 100$
  • B
    $a (100) > 100$
  • C
    $a (200) ≤ 100$
  • D
    None of these
Answer
Correct option: A.
$a(100) ≤ 100$
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MCQ 331 Mark
For each $n \in N,$ the correct statement is
  • A
    $2n < n$
  • B
    $nn2 > 2$
  • ✓
    $n4n < 10$
  • D
    $2713n > n +$
Answer
Correct option: C.
$n4n < 10$
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MCQ 341 Mark
Let $P(n) = 5^n− 2^n. P(n)$ is divisible by $3\lambda $ where $\lambda$ and $n$ both are odd positive integers, then the least value of $n$ and $\lambda$ will be:
  • A
    $13$
  • B
    $11$
  • ✓
    $1$
  • D
    $5$
Answer
Correct option: C.
$1$
$5^n− 2^n$ is divisible by $5 − 2 = 3$ always$...$
Putting $n = \lambda = 1$ which is the least odd positive integer,
​​​​​​​this works to be true.
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MCQ 351 Mark
$(2 ∙ 7^N+ 3 ∙ 5^N– 5)$ is divisible by $………..$ for all $N \in N:$
  • A
    $6$
  • B
    $12$
  • C
    $18$
  • ✓
    $24$
Answer
Correct option: D.
$24$
Let $P(n): (2 ∙ 7^n+ 3 ∙ 5^n– 5)$ is divisible by $24.$
For $n = 1,$ the given expression becomes $(2 ∙ 7^1+ 3 ∙ 5^1– 5) = 24$,
which is clearly divisible by $24.$
So, the given statement is true for $n = 1$,
i.e., $P(1)$ is true.
Let $P(k)$ be true. Then,
$ P(k):\left(2 \cdot 7^n+3 \cdot 5^n-5\right)$ is divisible by $24 $
$ \Rightarrow\left(2 \cdot 7^n+3 \cdot 5^n-5\right)=24 m$, for $m=N $
$ \text { Now, }\left(2 \cdot 7^n+3 \cdot 5^n-5\right) $
$ =\left(2 \cdot 7^k \cdot 7+3 \cdot 5^k \cdot 5-5\right) $
$ =7\left(2 \cdot 7^k+3 \cdot 5^k-5\right)$
$=6 \cdot 5^k+30 $
$ =(7 \times 24 m)-6\left(5^k-5\right) $
$ =(24 \times 7 m)-6 \times 4 p,$
where $\left(5^k-5\right)=5\left(5^{k-1}-1\right)=4 p$
$[$Since $\left(5^{k-1}-1\right)$ is divisible by $(5-1) ]$
$= 24 \times (7m – p)$
$= 24r,$ where $r = (7m – p) \in N$
$\Rightarrow P (k + 1): (2 ∙ 7^k+ 13 ∙ 5^k+ 1 – 5)$ is divisible by $24.$
$\Rightarrow P(k + 1)$ is true, whenever $P(k)$ is true.
Thus, $P(1)$ is true and $P(k + 1)$ is true, whenever $P(k)$ is true.
Hence, by the principle of mathematical induction, $P(n)$ is true for all $n \in N.$
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MCQ 361 Mark
Let $P(n) : “2^n< (1 \times 2 \times 3 \times … \times n)”$. Then the smallest positive integer for which $P(n)$ is true is:
  • A
    $1$
  • B
    $2$
  • C
    $3$
  • ✓
    $4$
Answer
Correct option: D.
$4$
$P(1) : 2^1 < 1$
$2< 1$ is false
$P(2) : 2^2 < 1 \times 2$
$4 < 2$ is false
$P(3) : 2^3 < 1 \times 2 \times 3$
$8 < 6$ is false
$P(4) : 2^4 < 1 \times 2 \times 3 \times 4$
$16 < 24$ is true
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MCQ 371 Mark
$x\left(x^{n-1}-n a^{n-1}\right) a^n(n-1)$ is divisible by $(x-a)^2$ for
  • A
    $n > 1$
  • B
    $n > 2$
  • ✓
    all $n \in N$
  • D
    None of the above
Answer
Correct option: C.
all $n \in N$
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MCQ 381 Mark
Consider the statement: $“P(n) : n^2 – n + 41$ is prime.” Then which one of the following is true?
  • ✓
    Both $P(3)$ and $P(5)$ are true.
  • B
    $P(3)$ is false but $P(5)$ is true.
  • C
    Both $P(3)$ and $P(5)$ are false.
  • D
    $P(5)$ is false but $P(3)$ is true.
Answer
Correct option: A.
Both $P(3)$ and $P(5)$ are true.
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MCQ 391 Mark
For all $\text{n}\in\text{N}\int\limits^\pi_0\frac{\text{sin(2}\text{n}\text{x)}}{\text{sin}\text{x}}\text{dx}$ is equal to:
  • ✓
    $0$
  • B
    $\pi$
  • C
    $\frac{\pi}{2}$
  • D
    $-\pi$
Answer
Correct option: A.
$0$
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MCQ 401 Mark
If $m, n$ are any two odd positive integer with $n < m,$ then the largest positive integers which divides all the numbers of the type $m^2– n^2$ is:
  • ✓
    $4$
  • B
    $6$
  • C
    $8$
  • D
    $9$
Answer
Correct option: A.
$4$
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MCQ 411 Mark
If $m, n$ are any two odd positive integer with $n< m,$ then the largest positive integers which divides all the numbers of the type $m^2- n^2$ is:
  • A
    $4$
  • B
    $6$
  • ✓
    $8$
  • D
    $9$
Answer
Correct option: C.
$8$
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MCQ 421 Mark
If $10n + 3. 4n + 2 k$ is divisible by $9$ for all $\ce{n Î N,}$ then the least positive integral value of $k$ is:
  • ✓
    $5$
  • B
    $3$
  • C
    $7$
  • D
    $1$
Answer
Correct option: A.
$5$
View full question & answer→
MCQ 431 Mark
Let $\ce{P(n) = n(n + 1)}$ is an even number, then which of the following satisfy $P(n):$
  • A
    $P(3)$
  • B
    $P(100)$
  • C
    $P(50)$
  • ✓
    All of these
Answer
Correct option: D.
All of these
Given, $\ce{P(n) = n(n + 1)}$
$\therefore P(3) = 3.4 = 12($even$)$
$P(100) = 100 \times 101 = 10100($even$)$
$P(50) = 50 \times 51 = 2520($even$)$
As $P(3), P(100), P(50)$ are even numbers.
Short cut Method: $\ce{P(n) = n(n + 1)}$
Product of two consecutive integer $($natural numbers$)$ is always even.
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MCQ 451 Mark
For principle of mathematical induction to be true, what type of number should $'n\ '$ be?
  • ✓
    Whole number
  • B
    Natural number
  • C
    Rational number
  • D
    Any form of number
Answer
Correct option: A.
Whole number
According to the Principle of Mathematical induction, $X(n)$ can be true if $X(1)$ is true and if $X(k)$ is true.
When $X(k)$ is true,
it implies that $X(k + 1)$ is also true.
Here $n$ can be equal to $0, 1, 2, 3$ and so on.
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MCQ 461 Mark
What is the sum of $13 + 23 + 33 + ........ + n^3$?
  • A
    $\Big(\frac{\text{n}(\text{n-1)}}{3}\Big)^2$
  • ✓
    $\Big(\frac{\text{n}(\text{n+1)}}{2}\Big)^2$
  • C
    $\Big(\frac{\text{n}(\text{n+1)}}{3}\Big)^2$
  • D
    $\Big(\frac{\text{n}(\text{n-1)}}{2}\Big)^2$
Answer
Correct option: B.
$\Big(\frac{\text{n}(\text{n+1)}}{2}\Big)^2$
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MCQ 471 Mark
If $P(k) = k^2(k + 3) (k^2– 1)$ is true, then what is $P(k + 1)\ ?$
  • A
    $ (k+1)^2(k+3)\left(k^2-1\right) $
  • B
    $ (k+1)^2(k+4)\left(k^2-1\right) $
  • ✓
    $ (k+1)^2(k+4) k(k+2) $
  • D
    $ (k+1)(k+4) k(k+2) $
Answer
Correct option: C.
$ (k+1)^2(k+4) k(k+2) $
In mathematical induction, if $P(k)$ is true,
we need to prove that $P(k + 1)$ is also true.
Here $P(k + 1)$ is found by substituting $(k + 1)$ in place of $k.$
$P(k + 1) = (k + 1)^2(k + 1 + 3) ((k + 1)^2– 1)$
$P(k + 1) = (k + 1)^2(k + 4) (k^2+ 1 + 2k – 1)$
$P(k + 1) = (k + 1)^2(k + 4) (k^2+ 2k)$
$P(k + 1) = (k + 1)^2(k + 4) k (k +2)$
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MCQ 481 Mark
For all $n \in N, 3n^5+ 5n^3+ 7n$ is divisible by:
  • A
    $5$
  • ✓
    $15$
  • C
    $10$
  • D
    $3$
Answer
Correct option: B.
$15$
Given number $= 3n^5+ 5n^3+ 7n$
Let $n = 1, 2, 3, 4, ……..$
$3n^5+ 5n^3+ 7n$
$= 3 \times 1^2+ 5 \times 1^3+ 7 \times 1$
$= 3 + 5 + 7 $
$= 15$
$3n^5+ 5n^3+ 7n$
$= 3 \times 2^5+ 5 \times 2^3+ 7 \times 2$
$= 3 \times 32 + 5 \times 8 + 7 \times 2$
$= 96 + 40 + 14$
$= 15 \times 10$
$= 150$
$3n^5+ 5n^3+ 7n$
$= 3 \times 3^5+ 5 \times 3^3+ 7 \times 3$
$= 3 \times 243 + 5 \times 27 + 7 \times 3$
$= 729 + 135 + 21$
$= 15 \times 59$
​​​​​​​$= 885$
Since, all these numbers are divisible by $15$ for $n = 1, 2, 3, …..$
So, the given number is divisible by $1$
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MCQ 491 Mark
If $n \in N,$ then $n(n^2− 1)$ is divisible by:
  • ✓
    $6$
  • B
    $16$
  • C
    $26$
  • D
    $24$
Answer
Correct option: A.
$6$
$\ce{n(n^2 − 1)= n(n − 1)(n + 1)}$
One of the $\ce{n, n + 1}$ and $n − 1$ will be a multiple of $3.$
Since $\ce{n − 1, n}$ and $n + 1$ are three consecutive integers,
therefore at least one of them will be divisible by $2.$
Therefore $\ce{n(n^2 − 1)}$ is divisible by $6.$
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MCQ 501 Mark
For every positive integer $n, n7/7 + n5/5 + 2n^3/3 - n/105$ is:
  • ✓
    an integer
  • B
    a rational number
  • C
    a negative real number
  • D
    an odd integer
Answer
Correct option: A.
an integer
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MCQ 511 Mark
$\ce{x(x^{n-1}- nα^{n-1}) + α^n(n - 1)}$ is divisible by $\ce{(x - α)^2}$ for:
  • A
    $n > 1$
  • B
    $n > 2$
  • ✓
    all $n \in N$
  • D
    None of the above
Answer
Correct option: C.
all $n \in N$
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MCQ 521 Mark
Let $T(k)$ be the statement $1 + 3 + 5 + .... + (2k – 1)= k +10$ Which of the following is correct:
  • A
    $T(1)$ is true
  • ✓
    $T(k)$ is true $\Rightarrow T(k + 1)$ is true
  • C
    $T(n)$ is true for all $n \in N$
  • D
    All above are correct
Answer
Correct option: B.
$T(k)$ is true $\Rightarrow T(k + 1)$ is true
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MCQ 531 Mark
The number of values of $n,$ for which $p(n) = 1! + 2! + 3! + 4!+ ...... + n!$ is the square of a natural number, is equal to:
  • A
    $0$
  • B
    $6$
  • C
    $2$
  • ✓
    $3$
Answer
Correct option: D.
$3$
For $n = 4, P(n) = 1 + 2 + 6 + 24 = 33.$
For $n > 4, n!$ will always have $0$ in the units place.
$3$ will be unit place digit of $p(n)$
hence can not be sqaure of any natural no.
So the $n = 1, 3$ are the answer.
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MCQ 541 Mark
If $n \in N,$ then $11^{n+2}+ 12^{2n+1}$ is divisible by:
  • A
    $113$
  • B
    $123$
  • ✓
    $133$
  • D
    None of these
Answer
Correct option: C.
$133$
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MCQ 551 Mark
$1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + n(n + 1):$
  • A
    $n(n + 1)(n + 2)$
  • B
    $\{n(n + 1)(n + 2)\}/2$
  • ✓
    $\{n(n + 1)(n + 2)\}/3$
  • D
    $\{n(n + 1)(n + 2)\}/4$
Answer
Correct option: C.
$\{n(n + 1)(n + 2)\}/3$
Let the given statement be $P(n).$ Then,
$P(n): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + n(n + 1) = (1/3)\{n(n + 1) (n + 2)\}$
Thus, the given statement is true for $n = 1,$
i.e., $P(1)$ is true.
Let $P(k)$ be true. Then,
$P(k): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + k(k + 1) = (1/3)\{k(k + 1) (k + 2)\}.$
Now, $1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +…+ k(k + 1) + (k + 1) (k + 2)$
$= (1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ……. + k(k + 1)) + (k + 1) (k + 2)$
$= (1/3) k(k + 1) (k + 2) + (k + 1)(k + 2) [$using $(i)]$
$= (1/3) [k(k + 1) (k + 2) + 3(k + 1)(k + 2)$
$= (1/3)\{(k + 1) (k + 2)(k + 3)\}$
$\Rightarrow P(k + 1): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +……+ (k + 1) (k + 2)$
$= (1/3)\{k + 1 )(k + 2) (k +3)\}$
$\Rightarrow P(k + 1)$ is true, whenever $P(k)$ is true.
Thus, $P(1)$ is true and $P(k + 1)$ is true, whenever $P(k)$ is true.
Hence, by the principle of mathematical induction, $P(n)$ is true for all values of $\in N$
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MCQ 561 Mark
$\forall n \in N; x^{2n-1}+ y^{2n-1}$ is divisible by?
  • A
    $x − y$
  • ✓
    $x + y$
  • C
    $xy$
  • D
    $x^2+ y^2$
Answer
Correct option: B.
$x + y$
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MCQ 571 Mark
If $P(n): “46 + 16 + k$ is divisible by $64$ for $\ce{n Î N”}$ is true, then the least negative integral value of $k$ is:
  • ✓
    $-1$
  • B
    $1$
  • C
    $2$
  • D
    $-2$
Answer
Correct option: A.
$-1$
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MCQ 601 Mark
Let $x > -1,$ then statement $p(n) : (1+x)^n > 1 + nx $, where $n \in N$ is true for
  • A
    For all $n \in N.$
  • B
    For all $n > 1.$
  • ✓
    For all $n > 1,$ provided $x = 0.$
  • D
    For all $n > 2.$
Answer
Correct option: C.
For all $n > 1,$ provided $x = 0.$
$p(1):(1+x)^1>1+x$ is false
$p(2):(1+x)^2>1+2 x \Rightarrow x^2>0$ is true when $x=0$
$p(3):(1+x)^3>1+3 x \Rightarrow x^2>0$ is true when $x=0$
Let $p(k)(1+x)^k>1+k x$ is true for some $k \in N, k>1$
$ \Rightarrow(1+x)^{k+1}>(1+k x)(1+x)$
$ \Rightarrow(1+x)^{k+1}>1+(k+1) x+k x^2>1+(k+1) x, x=0$
$\Rightarrow p (k+1)$ is true whenever $p(k)$ is true
so by principle of mathematical induction $p(n)$ is true for all $n > 1$ provided $x = 0$
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MCQ 611 Mark
$P(n): 2 \times 7n + 3 \times 5n - 5$ is divisible by:
  • ✓
    $24, \forall n \in N$
  • B
    $21, \forall n \in N$
  • C
    $32, \forall n \in N$
  • D
    $50, \forall n \in N$
Answer
Correct option: A.
$24, \forall n \in N$
Concepts:
Suppose there is a given statement $P(n)$ involving the natural number $n$ such that
The statement is true for $n = 1,$
i.e., $P(1)$ is true, and
If the statement is true for $n = k ($where $k$ is some positive integer$)$, then the statement is also true for $n = k + 1,$
i.e., the truth of $P(k)$ implies the truth of $P(k + 1).$
Then, $P(n)$ is true for all natural numbers $n$​​​​​​​
Calculation:
Given:
$P(n) = 2 \times 7n + 3 \times 5n - 5$
Put $n = 1$
$P(1) = 2 \times 7^1+ 3 \times 5^1- 5 = 24$, Which is divisible by $24$
Assume $P(k)$ is true
$P(k) = 2 \times 7^k+ 3 \times 5^k- 5 = 24q$, where $q\ \epsilon\ N .....(1)$
Now,
$ T(k+1)=2 \times 7^{k+1}+3 \times 5^{k+1}-5=2 \times 7^k \times 7+3 \times 5^k \times 5-5 $
$ \Rightarrow 7\left\{2 \times 7^k+3 \times 5^k-5-3 \times 5^k+5\right\}+3 \times 5^k \times 5-5 $
$ \Rightarrow 7\left\{24 q-3 \times 5^k+5\right\}+15 \times 5^k-5 $
$ \Rightarrow(7 \times 24 q)-21 \times 5^k+35+15 \times 5^k-5 $
$ \Rightarrow(7 \times 24 q)-6 \times 5^k+30=(7 \times 24 q)-6\left(5^k-5\right) $
$\Rightarrow(7 \times 24 q)-6 \times(4 p)$ As $(5^k-5) $ is a multiple of $4 \}$
$\Rightarrow (7 \times 24q) - 24p = 24(7q - p)$
$\Rightarrow 24 \times r, r = 7q - p,$ is some natural number $.......(2)$
Thus, $P(k + 1)$ is true whenever $P(k)$ is true
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MCQ 621 Mark
For each natural number, the statement $P(n) = 2^{3n}− 1$ is divisible by:
  • A
    $10$
  • B
    $6$
  • ✓
    $7$
  • D
    None of these.
Answer
Correct option: C.
$7$
$ P(n)=2^{3 n}-1=8^n-1 $
$ P(n)=(1+7)^n-1 $
$ \Rightarrow P(n)=1+{ }^n C_1 \times 7+{ }^n C_2 \times 7^2+\ldots+{ }^n C_n \times 7^n-1 $
$ \Rightarrow P(n)=7\left({ }^n C_1+{ }^n C_2 7+\ldots+{ }^n C_n 7{ }^{n-1}\right)$
Therefore, $P(n)$ is divisible by $7.$
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MCQ 631 Mark
Consider the following statements : $1) \sim (p\ ∧ q) = \sim p\ ∨ \sim q 2) \sim (p\ ∨ q) = \sim p\ ∧ \sim q 3) \sim (\sim p) = p$ Which of the above statements is/are correct?
  • A
    $1$ and $2$
  • B
    $2$ and $3$
  • ✓
    $1, 2$ and $3$
  • D
    None of these
Answer
Correct option: C.
$1, 2$ and $3$
Concept:
The negation of a conjunction $p ∧ q$ is the disjunction of the negation of $p$ and the negation of $q.$ Equivalently, we write $\sim (p ∧ q) = \sim p ∨ \sim q$
The negation of a disjunction $p ∨ q$ is the conjunction of the negation of $p$ and the negation of $q.$ Equivalently, we write $\sim (p ∨ q) = \sim p ∧ \sim q$
Negation of negation of a statement is the statement itself. Equivalently, we write $\sim (\sim p) = p$
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MCQ 641 Mark
For each $n\ \epsilon\ N,$ then $3^{2n+1}+ 1$ is divisible by:
  • ✓
    $2$
  • B
    $3$
  • C
    $7$
  • D
    None of these
Answer
Correct option: A.
$2$
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MCQ 651 Mark
If $n \in N \Big(\frac{\text{n}+1}{2}\Big)^\text{n}\geq\text{n!}$ is true when:
  • ✓
    $n ≥ 1$
  • B
    $n ≥ 2$
  • C
    $n > 1$
  • D
    $n > 2$
Answer
Correct option: A.
$n ≥ 1$
Concept:
$\text{n!}=\text{n}\times(\text{n}-1)\times\text{n}-2.....\times3\times2\times1$
Calculation:
Given:
$\text{p}\text{(n)}=\Big(\frac{\text{n+1}}{2}\Big)^\text{n}\geq\text{n!}$
Put $n = 1$
$\text{p}(2)=\Big(\frac{2+1}{2}\Big)^2\geq2!$
$=\Big(\frac{3}{2}\Big)^2\geq2\times1$
$2.25 ≥ 2$
Put $n = 3$
$\text{p}(3)=\Big(\frac{3+1}{2}\Big)^3\geq3!$
$8\geq3\times2\times1,8\geq6$
Hence, The given expression $P(n)$ is true for $n ≥ 1$
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MCQ 661 Mark
For all positive integral values ofn, $3^{2n} - 2n + 1$ is divisible by:
  • ✓
    $2$
  • B
    $4$
  • C
    $8$
  • D
    $12$
Answer
Correct option: A.
$2$
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MCQ 671 Mark
If $P(n)$ be the statement $n(n + 1) + 1$ is odd, then which of the following is false?
  • A
    $P(2)$
  • B
    $P(3)$
  • C
    $P(4)$
  • ✓
    None of these.
Answer
Correct option: D.
None of these.
$P(n) = n(n + 1) + 1$
$P(2) = 6 + 1 = 7$
$P(3) = 3 \times 4 + 1 = 13$
$P(4) = 4 \times 5 + 1 = 21$
None of the above is even.
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MCQ 681 Mark
$n(n + 1)(n + 5)$ is a multiple of $3$ is true for:
  • A
    All natural numbers $n>5$
  • B
    Only natural number $3 ≤ n < 15$
  • ✓
    All natural numbers $n$
  • D
    None
Answer
Correct option: C.
All natural numbers $n$
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MCQ 691 Mark
If $P(n): “49^n+ 16^n+ k$ is divisible by $64$ for $n \in N”$ is true, then the least negative integral value of $k$ is:
  • A
    $1$
  • B
    $2$
  • ✓
    $-1$
  • D
    $-3$
Answer
Correct option: C.
$-1$
Given that $P(n): 49^n+ 16^n+ k$ is divisible by $64$ for $n \in N$
For $n = 1,$
$P(1): 49 + 16 + k = 65 + k$ is divisible by $64.$
Thus $k,$ should be $-1$ since, $65 - 1 = 64$ is divisible by $64.$
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MCQ 701 Mark
For all $n \in N, 3.5^{2n+1}+ 2^{3n+1}$ is divisible by:
  • A
    $19$
  • ✓
    $17$
  • C
    $23$
  • D
    $25$
Answer
Correct option: B.
$17$
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MCQ 711 Mark
If $n$ is an odd positive integer, then $a^n + b^n$ is divisible by:
  • A
    $a^2 + b^2$
  • ✓
    $a + b$
  • C
    $a – b$
  • D
    none of these
Answer
Correct option: B.
$a + b$
Given number $= a^n + b^n$
Let $n = 1, 3, 5, ……..$
$a^n + b^n = a + b$
$a^n + b^n = a^3 + b^3 = (a + b) \times (a^2 + b^2 + ab)$ and so on.
Since, all these numbers are divisible by $(a + b)$ for $n = 1, 3, 5,…..$
So, the given number is divisible by $(a + b)$
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MCQ 731 Mark
If $10^\text{n} + 3 \times 4^{\text{n}+2}+\lambda$ is divisible by $9$ for all $\text{n}\in\text{N},$ then the least positive integer value of $\lambda$ is
  • ✓
    $5$
  • B
    $3$
  • C
    $7$
  • D
    $1$
Answer
Correct option: A.
$5$
Given,
$10\text{n }+3\times^{\text{n}+2}+\lambda$ is divisible by $9,$
$\text{P}(1)=10^1+3\times4^{1+2}+\lambda$ is exactly divisible by $9$ then the value of $\lambda$ is $5.$
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MCQ 741 Mark
For all $n\in N, 72n − 48n−1$ is divisible by:
  • A
    $50$
  • ✓
    $2304$
  • C
    $1234$
  • D
    $44$
Answer
Correct option: B.
$2304$
Concepts:
Suppose there is a given statement $P (n)$ involving the natural number $n$ such that
The statement is true for $n = 1,$
i.e., $P (1)$ is true, and
If the statement is true for $n = k ($where $k$ is some positive integer$)$, then the statement is also true for $n = k + 1,$ i.e., truth of $P (k)$ implies the truth of $P (k + 1).$
Then, $P (n)$ is true for all natural numbers $n$
Calculation:
Given:
$P(n) = 72n − 48n−1$
Put, $n = 1$
$P(1) = 72 − 48 \times 1 −1 = 0$
Check the expression $P(n)$ for $n = k ($where $k$ is some positive integer$) = 2, 3, 4......$
$P(2) = 7^{2n}− 48n − 1$
$= 7^4− 48 \times 2 − 1$
$= 2401 – 96 – 1$
$= 2401 – 97$
$= 2304$
$P(3) = 7^{2n}− 48n − 1$
$= 7^6− 48 \times 3 − 1$
$= 117649 – 144 – 1$
$= 117649 – 145$
$= 117504$
$= 2304 \times 51$
Since, all these numbers are divisible by $2304$ for $n = 1$ and $k = 2, 3,…..$
So, the given number is divisible by $2304$
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MCQ 751 Mark
Let $P(n)$ be the statement that $n^2 - n + 41$ is prime, then which of the following is not true?
  • A
    $P(2)$
  • B
    $P(3)$
  • ✓
    $P(41)$
  • D
    none of these
Answer
Correct option: C.
$P(41)$
given that $P(n) = n^2- n + 41$
$P(2) = 2^2- 2 + 41 = 43 ($prime true$)$
$P(3) = 3^2- 3 + 41 = 47 ($prime true$)$
$P(41) = 41^2- 41 + 41 = (41)^2$
$\therefore P(41 )$ is not true.
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MCQ 761 Mark
What is the sum of $2 + 4 + 6 + 8 +....+ 2n\ ?\ A:$
  • ✓
    $n(n + 1)$
  • B
    $n(n + 2)$
  • C
    $n(n + 3)$
  • D
    $n(n + 4)$
Answer
Correct option: A.
$n(n + 1)$
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MCQ 771 Mark
If $x^n− 1$ is divisible by $x − k,$ then the least positive integral value of $k$ is:
  • ✓
    $1$
  • B
    $2$
  • C
    $3$
  • D
    $4$
Answer
Correct option: A.
$1$
if $f(x) = x^n− 1$ is divisible by $x − k$
Then $f(k) = 0$
Therefore, $k^n= 1$
and thus least positive integral value of $k$ is $1$
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MCQ 781 Mark
The smallest positive integer $n$ for which $ \text{n}!<\Big(\frac{\text{n}+1}{2}\Big)\text{n}$ holds, is:
  • A
    $1$
  • ✓
    $2$
  • C
    $3$
  • D
    $4$
Answer
Correct option: B.
$2$
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MCQ 791 Mark
$72n + 16n - 1$ is divisible by $(n \in N):$
  • ✓
    $64$
  • B
    $49$
  • C
    $13$
  • D
    $25$
Answer
Correct option: A.
$64$
$S = 72n + 16n - 1$
For $n = 1$
$S = 49 + 16 - 1 = 64$
For $n = 2$
$S = 2401 + 32 - 1$
$= 2432$
$= 64 \times 32,$
which is divisible by $64$
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MCQ 801 Mark
What would be the hypothesis of mathematical induction for $n(n + 1) < n! ($where $n ≥ 4)\ ?$
  • A
    It is assumed that at $n = k, k(k + 1)! > k!$
  • ✓
    It is assumed that at $n = k, k(k + 1)! < k!$
  • C
    It is assumed that at $n = k, k(k + 1)! > (k + 1)!$
  • D
    It is assumed that at $n = k, (k + 1)(k + 2)! < k!$
Answer
Correct option: B.
It is assumed that at $n = k, k(k + 1)! < k!$
When we use the principle of mathematical induction, we assume that $P(n)$ is true for $P(k)$ and prove that $P(k + 1)$ is also true.
Here $P(k)$ is $k(k + 1)! < k!$
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MCQ 811 Mark
If $P(n): 3n < n!, n\ \epsilon\ N,$ Then $P(n)$ is true for:
  • ✓
    $n ≥ 7$
  • B
    $n ≥ 3$
  • C
    $n ≥ 6$
  • D
    all 
Answer
Correct option: A.
$n ≥ 7$
Concept:
$\text{n}!=\text{n}\times(\text{n}-1)\times(\text{n}-2).....\times3\times2\times1$
Given:
$P(n): 3n < n!$
This can be solved directly by hit and trial method, putting the option in expression and checking its validity
We will choose first the smallest number from the options
Putting $n = 3$
$ P(n) = 3n < n! = 3^3 < 3! $
$\Rightarrow 27 ≮ 3 \times 2 \times 1, 27 ≮ 6 ,$
Hence Option $2$ is wrong
Putting $n = 6$
$P(n) = 3n < n! = 63 < 6!$
$P(n) = 3n < n! = 3^6 < 6! , 1029 < 6 \times 5 \times 4 \times 3 \times 2 \times 1$
$1029 ≮ 720$ Hence Option $3$ is wrong
Put $n = 7$
$P(n) = 3n < n! = 3^7 < 7! , 2187 < 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
$2187 < 5040,$ Option $1$ satisfies the given expression,
Hence the correct answer is option $1.$
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MCQ 821 Mark
For all $n\in N, 5^{2n}− 1$ is divisible by:
  • A
    $26$
  • ✓
    $24$
  • C
    $11$
  • D
    $25$
Answer
Correct option: B.
$24$
Given number $= 5^{2n}− 1$
Let $n = 1, 2, 3, 4, ……..$
$ 5^{2 n}-1=5^2-1=25-1=24 $
$ 5^{2 n}-1=5^4-1=625-1=624=24 \times 26 $
$ 5^{2 n}-1=5^6-1=15625-1=15624=651 \times 24 $
Since, all these numbers are divisible by $24$ for $n = 1, 2, 3, …..$
So, the given number is divisible by $2$
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MCQ 831 Mark
For a positive integer $n,$ let $\text{a(n)} =1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+…+\frac{1}{\text{n.}} $ Then,
  • ✓
    $a(100) ≤ 100$
  • B
    $a(100) >100$
  • C
    $a(200) ≤ 100$
  • D
    None of these
Answer
Correct option: A.
$a(100) ≤ 100$
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MCQ 841 Mark
A student was asked to prove a statement $P(n)$ by induction. He proved that $P(k + 1)$ is true whenever $P(k)$ is true for all $k > 5 \in N$ and also that $P(5)$ is true. Based on this, he could conclude that $P(n)$ is true:
  • A
    for all $n \in N$
  • B
    for all $n > 5$
  • ✓
    for all $n ≥ 5$
  • D
    for all $n < 5$
Answer
Correct option: C.
for all $n ≥ 5$
The student could be able to conclude that $P(n)$ is true for all $n ≥ 5$ since $P(5)$ is true for all $k > 5 \in N$ as well as true for $P(5)$ and $P(k + 1)$ is true, whenever $P(k)$ is true.
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MCQ 851 Mark
For positive integer $n, 10^{n-2} > 81 n,$ if:
  • A
    $n > 5$
  • ✓
    $n ≥ 5$
  • C
    $n < 5$
  • D
    $n > 6$
Answer
Correct option: B.
$n ≥ 5$
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MCQ 861 Mark
Choose the correct answer. If $x^n- 1$ is divisible by $x - k,$ then the least positive integral value of $k$ is:
  • ✓
    $1$
  • B
    $2$
  • C
    $3$
  • D
    $4$
Answer
Correct option: A.
$1$
Let $P(n) = x^n- 1$ is divisible by $x - k$
$P(1) = x - 1$ is divisible by $x - k.$
Since $x - 1$ is divisible by $x - 1,$ the least integral value of $k$ is $1.$
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MCQ 871 Mark
$n (n+1) (n+5)$ is a multiple of $3$ is true for
  • A
    All natural numbers $n > 5$
  • B
    Only natural number $3 ≤ n < 15$
  • ✓
    All natural numbers $n$
  • D
    None
Answer
Correct option: C.
All natural numbers $n$
Let the statement be denoted by $p(n)$
i.e., $P(n) : n (n+1) (n+5)$ is a multiple of $3$
For $n = 1, n(n+1) (n+5) = 1.2.6 = 12 = 3.4$
$P(n)$ is true for $n = 1$
Suppose $p(k)$ is true for $n = k$
i.e. $k(k+1) (k+5) =3m ($let$)$ or $k^3 + 6k^2 + 5k = 3m ... (i)$
Replacing $k$ by $k+1,$ we get
$(k+1) (k+2) (k+6) = k (k^2+ 8k + 12) + (k^2+ 8k + 12) $
$k^3+ 9k^2+ 20k + 12 = (k^3+ 6k^2+ 5k) + (3k^2+15k+12)$
$= 3m + 3k^2+ 15k + 12 [$from $(i)]$
$= 3 (m + k^2+ 5k + 4)$
i.e. $(k+1) (k+2) (k+6)$ is a multiple of $3$
i.e. $P(k+1)$ is multiple of $3,$ if $P(k)$ is a multiple of $3$
i.e. $P(k+1)$ is true whenever $P(k)$ is true.
Hence $P(n)$ is true for all $n \in N$
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MCQ 881 Mark
The greatest positive integer, which divides $n(n +1)(n + 2)(n + 3)$ for all $\text{n Î N},$ is:
  • A
    $2$
  • B
    $6$
  • ✓
    $24$
  • D
    $120$
Answer
Correct option: C.
$24$
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MCQ 901 Mark
If an $= \sqrt7 + \sqrt7 + \sqrt7 +......$ having n radical signs then by methods of mathematical induction which is true:
  • A
    $\text{an}>7"\text{n}\geq1$
  • ✓
    $\text{an}<7"\text{n}\geq1$
  • C
    $\text{an}<4"\text{n}\geq1$
  • D
    $\text{an}<3"\text{n}\geq1$
Answer
Correct option: B.
$\text{an}<7"\text{n}\geq1$
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MCQ 911 Mark
The sum of the series $1^3 + 2^3 + 3^3 + ………..n^3$ is:
  • A
    $\Big(\frac{\text{(n}+1)}{2}\Big)^2$
  • B
    $\Big(\frac{\text{n}}{2}\Big)^2$
  • C
    $\Big(\frac{\text{n}\text{(n}+1)}{2}\Big)$
  • ✓
    $\Big(\frac{\text{n}\text{(n}+1)}{2}\Big)^2$
Answer
Correct option: D.
$\Big(\frac{\text{n}\text{(n}+1)}{2}\Big)^2$
Given, series is $1^3 + 2^3 + 3^3 + ……….. n^3$
$\Big(\frac{\text{n}\text{(n}+1)}{2}\Big)^2$
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MCQ 921 Mark
Choose the correct answer. For all $n \in N, 3 \times 5^{2n+1} + 2^{3n+1}$ is divisible by:
  • A
    $19$
  • B
    $17$
  • ✓
    $23$
  • D
    $25$
Answer
Correct option: C.
$23$
$23$
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MCQ 931 Mark
For $n\in N,\big(\frac{1}{5}\big)\text{n}^5+\big(\frac{1}{3}\big)\text{n}^3+\big(\frac{1}{15}\big)$ is:
  • A
    an integer
  • ✓
    a natural number
  • C
    a positive fraction
  • D
    None of these
Answer
Correct option: B.
a natural number
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MCQ 941 Mark
$P(n) = n(n^2– 1)$. Which of the following does not divide $P(k+1)\ ?$
  • A
    $k$
  • B
    $k + 2$
  • ✓
    $k + 3$
  • D
    $k + 1$
Answer
Correct option: C.
$k + 3$
$P(n) = n(n^2- 1)$
$P(k + 1) = (k + 1) ((k + 1)^2-1)$
$P(k + 1) = (k + 1) (k^2+ 1 + 2k - 1)$
$P(k + 1) = (k + 1) (k^2+ 2k)$
$P(k + 1) = (k + 1) k (k + 2)$
Therefore, $k, (k + 1), (k - 1)$ divide $P(k + 1)$.
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MCQ 951 Mark
The greatest positive integer, which divides $(n + 2) (n + 3) (n + 4) (n + 5) (n + 6)$ for all $n \in N,$ is:
  • A
    $4$
  • ✓
    $120$
  • C
    $240$
  • D
    $24$
Answer
Correct option: B.
$120$
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MCQ 961 Mark
$n^2 < 2^n$ for all natural numbers:
  • ✓
    $n ≥ 5$
  • B
    $n < 5$
  • C
    $n > 1$
  • D
    $n ≤ 3$
Answer
Correct option: A.
$n ≥ 5$
Consider, $P(n) : n^2 < 2^n$
Substituting $n = 1, 2, 3,…$
$P(1): 1^2< 2^1$
$1 < 2 ($not true$)$
$P(2): 2^2 < 2^2$
$4 < 4 ($not true$)$
$P(3): 3^2 < 2^3$
$9 < 8 ($not true$)$
$P(4): 4^2 < 2^4$
$16 < 16 ($not true$)$
$P(5): 5^2 < 2^5$
$25 < 32 ($true$)$
$P(6): 6^2 < 2^6$
$26 < 64 ($true$)$
Thus, $n^2 < 2^n$ for all natural numbers $n ≥ 5.$
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MCQ 971 Mark
If $P(n)$ is a statement such that $P(3)$ is true. Assuming $P(k)$ is true $Þ P(k + 1)$ is true for all $\text{k} \geq 3$, then $P(n)$ is true:
  • A
    for all $n$
  • ✓
    for $n ≥ 3$
  • C
    for $n > 4$
  • D
    None of these
Answer
Correct option: B.
for $n ≥ 3$
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MCQ 981 Mark
Let $P(n)$ be a statement and $P(n) = P(n + 1)\forall n \in N,$ then $P(n)$ is true for what values of $n?$
  • ✓
    For all $n$
  • B
    For all $n > 1$
  • C
    For all $n > m , m$ being a fixed positive integer
  • D
    Nothing can be said
Answer
Correct option: A.
For all $n$
Given, $P(n) = P(n+1)\forall n \in N$
Substituting $n - 1$ in place of $n,$
$P(n - 1) = P(n)$
Thus if $P(k)$ is true for some $k \in N,$ then it is true for $k - 1$ and $k + 1.$
Thus, it is true $\forall k \in N$
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MCQ 991 Mark
$x\left(x^{n-1}-n a^{n-1}\right)+a^n(n-1)$ is divisible by $(x-a)^2$ for:
  • A
    $n > 1$
  • B
    $n > 2$
  • ✓
    all $n \in N$
  • D
    None of the above
Answer
Correct option: C.
all $n \in N$
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MCQ 1001 Mark
A student was asked to prove a statement $p(n)$ by induction. He proved $p(K + 1)$ is true whenever $p(k)$ is true for all $\text{k}>5\in\text{N}$ and also $p(5)$ is true. On the basis of this he could conclude that $p(n)$ is true.
  • A
    For all $\text{n}\in\text{N}$
  • B
    For all $n > 4$
  • ✓
    For all $\text{n}\geq5$
  • D
    For all $n > 5$
Answer
Correct option: C.
For all $\text{n}\geq5$
$P(n)$ is true for all positive integer $n,$
i.e. $\text{n}\geq5,$
Where $P(n)$ is a Propositional function, complete two steps:
Basic Step: Verify that the proposition $P(1)$ is true.
Inductive Step: Show the conditional statement,
$\big[\text{P}(1) \wedge \text{P}(2) \wedge-\wedge\text{P}(\text{k})\big]\rightarrow\text{P(k+1)}$ holds for all positive integer.
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MCQ 1011 Mark
For all $\text{n Î N,} 3.52n + 1 + 23n + 1$ is divisible by:
  • A
    $19$
  • ✓
    $17$
  • C
    $23$
  • D
    $25$
Answer
Correct option: B.
$17$
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MCQ 1021 Mark
$2.4^{2\text{n+1}}+3^{3\text{n+1}}$ is divisible by: $($for all $n \in N)$
  • A
    $2$
  • B
    $9$
  • C
    $3$
  • ✓
    $11$
Answer
Correct option: D.
$11$
Concepts:
Suppose there is a given statement $p(n)$ involving the natural number $n$ such that
The statement is true for $n = 1,$
i.e., $P(1)$ is true, and
If the statement is true for $n = k ($where $k$ is some positive integer$)$, then the statement is also true for $n = k + 1,$
i.e., truth of $P (k)$ implies the truth of $P(k + 1).$
Then, $P (n)$ is true for all natural numbers $n.$​​​​​​​
Calculation:
Given:
$\text{p}\text{(n)}=2.4^{2\text{n+1}}+3^{3\text{n+1}}$
Take $n = 1$
$\text{p}(1)=2.4^{2\times+1}+3^{3\times+1}$
$=2.4^3+3^4=209=11\times19$
Therefore we can say that $P (n)$ is divisible by $11.$
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MCQ 1031 Mark
Let $P(n)$ denotes the statement that $n n^2 +$ is odd. It is seen that $P(n) Þ P(n + 1), P(n)$ is true for all:
  • A
    $n > 1$
  • B
    $n$
  • C
    $n > 2$
  • ✓
    None of these
Answer
Correct option: D.
None of these
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MCQ 1041 Mark
If $x^n- 1$ is divisible by $\text{x}-\lambda,$ then the least positive integral value of $\lambda$ is:
  • ✓
    $1$
  • B
    $2$
  • C
    $3$
  • D
    $4$
Answer
Correct option: A.
$1$
Given
$x^n- 1$
We know that
$x = k$ is the root of the equation $(x - 1)$
$\Rightarrow x^n- 1 = 0$
$\Rightarrow x^n= 1$
Hence, the least positive integral value of $\lambda$ is $1.$
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MCQ 1051 Mark
By mathamatical induction $n(n^2− 1)$ is divisible by:
  • A
    $19$
  • B
    $23$
  • ✓
    $24$
  • D
    $29$
Answer
Correct option: C.
$24$
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MCQ 1061 Mark
For every positive integer n, $\text{x},\frac{\text{n}^7}{7}+\frac{\text{n}^6}{5}+\frac{2\text{n}^3}{3}-\frac{\text{x}}{105}$ is:
  • A
    an integer
  • ✓
    a rational number
  • C
    a negative real number
  • D
    an odd integer
Answer
Correct option: B.
a rational number
b. a rational number
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MCQ 1071 Mark
$n(n + 1) (n + 5)$ is a multiple of:
  • ✓
    $3$
  • B
    $8$
  • C
    $5$
  • D
    $7$
Answer
Correct option: A.
$3$
Let $P(n) = n(n + 1)(n + 5)$
Substituting $n = 1, 2, 3,….$
$P(1) = 1(1 + 1)(1 + 5) = 2(6) = 12;$ multiple of $2, 3, 4, 6$
$P(2) = 2(2 + 1)(2 + 5) = 2(3)(7) = 42;$ multiple of $2, 3, 6, 7$
$P(3) = 3(3 + 1)(3 + 5) = 3(4)(8) = 96;$ multiple of $2, 3, 4, 6, 8, 12$
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MCQ 1081 Mark
If $n \in N,$ then $121n - 25n + 1900n - (-4) n$ is divisible by which one of the following?
  • A
    $1904$
  • ✓
    $2000$
  • C
    $2002$
  • D
    $2006$
Answer
Correct option: B.
$2000$
Concepets:
Suppose there is a given statement $P (n)$ involving the natural number $n$ such that
The statement is true for $n = 1,$
i.e., $P (1)$ is true, and
If the statement is true for $n = k ($where $k$ is some positive integer$),$ then the statement is also true for $n = k + 1,$
i.e., truth of $P (k)$ implies the truth of $P (k + 1)$.
Then, $P (n)$ is true for all natural numbers $n$
Caluculation:
Given:
$P(n) = 121^n– 25^n+ 1900^n– (-4)^n$
Now, $P(1) = 121^1– 25^1+ 1900^1– (-4)^1$
$\Rightarrow P (1) = 121 – 25 + 1900 + 4$
$\Rightarrow P (1) = 2000$
Therefore we can say that $P (n)$ is divisible by $2000.$
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MCQ 1091 Mark
If $n \in N,$ then $11^{n+2}+ 12^{2n+1}$ is divisible by:
  • A
    $113$
  • B
    $123$
  • ✓
    $133$
  • D
    None of these
Answer
Correct option: C.
$133$
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MCQ 1101 Mark
Let us consider the series $S_n= 2.7^n+ 3.5^n- 5.$ If $S_n$ is divisible for every $n$, then $S_n$ is:
  • ✓
    $> 0$
  • B
    $> 1$
  • C
    $> 5$
  • D
    None of these
Answer
Correct option: A.
$> 0$
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MCQ 1111 Mark
Mathematical Induction is the principle containing the set.
  • A
    $R$
  • ✓
    $N$
  • C
    $Q$
  • D
    $Z$
Answer
Correct option: B.
$N$
Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers.
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MCQ 1121 Mark
For all $n \in N, 3^{2n} + 7$ is divisible by:
  • ✓
    $8$
  • B
    $3$
  • C
    $11$
  • D
    non of these
Answer
Correct option: A.
$8$
Given number $= 32n + 7$
Let $n = 1, 2, 3, 4, ……..$
$3^{2 n}+7=3^2+7=9+7=16$
$3^{2 n}+7=3^4+7=81+7=88$
$3^{2 n}+7=3^6+7=729+7=736$
Since, all these numbers are divisible by $8$ for $n = 1, 2, 3, …..$
So, the given number is divisible by $8$
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MCQ 1131 Mark
For every integer $\text{n}\geq1,$ $(3^{2n}-1)$ is always divisible by:
  • A
    $2^{n2}$
  • B
    $2^{n+4}$
  • ✓
    $2^{n+2}$
  • D
    $2^{n+3}$
Answer
Correct option: C.
$2^{n+2}$
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MCQ 1141 Mark
The sum of n terms of the series $1^2 + 3^2 + 5^2 +………$ is:
  • ✓
    $n(4n^2 – 1)/3$
  • B
    $n^2(2n^2 + 1)/6$
  • C
    none of these.
  • D
    $n^2(n^2 + 1)/3$
Answer
Correct option: A.
$n(4n^2 – 1)/3$
Let $S = 1^2 + 3^2 + 5^2 +………(2n – 1)^2$
$\Rightarrow S = {1^2 + 2^2 + 3^2 + 4^2 ………(2n – 1)^2 + (2n)^2} – {2^2 + 4^2 + 6^2 +………+ (2n)^2}$
$\Rightarrow S = {2n \times (2n + 1) \times (4n + 1)}/6 – {4n \times (n + 1) \times (2n + 1)}/6$
$\Rightarrow S = n(4n^2 – 1)/3$
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MCQ 1171 Mark
For every integer $\text{n} ≥ 1, ({3^2}^{\text{n}}-1)$ is always divisible by.
  • A
    $2^{\text{n}^2}$
  • B
    $2^{\text{n + 4}}$
  • ✓
    $2^{\text{n + 2}}$
  • D
    $2^{\text{n + 3}}$
Answer
Correct option: C.
$2^{\text{n + 2}}$
For $\text{n = 1},\ 3^{\text{2}^1}-1=8,$ which is divisible by $2^{\text{n + 2}}$
Let us assume that $3^{2^{\text{m}}}-1$ is divisible by $2^\text{m + 2}$ for some integral value of $m.$
Let us consider the expression for $m+1$
$3^{2^{\text{m+1}}}-1$
$=(3^{2^{\text{m}}}-1)\ \times ​​(3^{2^{\text{m}}}+1)$
The first term is divisible $2^{\text{m+2}}$ and the second term is also an even number.
Hence, the term is divisible by $2^{\text{m+2}}$
Hence, by induction we can prove that it is true for all $m.$
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MCQ 1181 Mark
For every positive integer $n, 7n – 3n$ is divisible by
  • A
    $2$
  • ✓
    $4$
  • C
    $5$
  • D
    $6$
Answer
Correct option: B.
$4$
Concept:
Suppose there is a given statement $P(n)$ involving the natural number $n$ such that
The statement is true for $n = 1,$
i.e., $P(1)$ is true, and
If the statement is true for $n = k ($where $k$ is some positive integer$)$, then the statement is also true for $n = k + 1,$ i.e., truth of $P(k)$ implies the truth of $P(k + 1).$
Then, $P(n)$ is true for all natural numbers $n$
Calculation:
We have to find $7^n- 3^n$ is divisible by which number
Consider $P(n): 7n - 3n$
$P (1): 7^1- 3^1= 4$
Thus, $7n - 3n$ is divisible by $4$
Let $P(k)$ is true for $n = K$
$\Rightarrow 7^k− 3^k$ is divisible by $4$
So, $7n – 3n = 4d$
Now, prove that $P(k+1)$ is true.
$\Rightarrow 7^{(k+1)}-3^{(k+1)}=7^{(k+1)}-7.3^k+7.3^k-3^{(k+1)} $
$ =7\left(7^k-3^k\right)+(7-3) 3^k $
$ =7(4 d)+(7-3) 3^k $
$ =7(4 d)+4.3^k $
$ =4\left(7 d+3^k\right) $
Hence, $P (n): 7^n- 3^n$ is divisible by $4$ is true.
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MCQ 1191 Mark
If $\forall m \in N,$ then $11^{m+2}+ 12^{2m-1}$ is divisible by:
  • A
    $121$
  • B
    $132$
  • C
    $133$
  • ✓
    None of these
Answer
Correct option: D.
None of these
View full question & answer→
MCQ 1201 Mark
$3 + 13 + 29 + 51 + 79 +...$ to $n$ terms $=:$
  • A
    $2 n^2+7 n^3$
  • B
    $n^2+5 n^3$
  • ✓
    $ n^3+2 n^2$
  • D
    none of these
Answer
Correct option: C.
$ n^3+2 n^2$
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MCQ 1211 Mark
For all positive integral values of n, $43^{2n}- 2n + 1$ is divisible by:
  • ✓
    $2$
  • B
    $4$
  • C
    $8$
  • D
    $12$
Answer
Correct option: A.
$2$
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MCQ 1221 Mark
$n^2+ 3n$ is always divisible by which number, provided $n$ is an integer?
  • ✓
    $2$
  • B
    $3$
  • C
    $4$
  • D
    $5$
Answer
Correct option: A.
$2$
$P(n) = n^2+ 3n$
$P(1) = 1 + 3$
$P(1) = 4$
Let’s assume that $P(k)$ is true and divisible by $4.$
Therefore, $P(k) = k^2+ 3k$ can be written as $4c.$
We need to check if $P(k + 1)$ is divisible by $4$
$P(k+1)=(k+1)^2+3(k+1) $
$P(k+1)=k^2+1+2 k+3 k+3 $
$P(k+1)=k^2+5 k+4 $
$P(k+1)=\left(k^2+3 k\right)+2 k+4 $
$P(k+1) = 4c + 2k + 4$
$P(k+1) = 4c + 2(k + 2)$
Clearly the second part of the equation is not divisible by $4.$
However $P(k) = 4c$ is divisible by $2$ and
$P(k + 1)$ is also divisible by $2.$
Therefore, $2$ divides $P(n)$
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MCQ 1231 Mark
If $x^{2n-1}+ y^{2n-1}$ is divisible by $x + y,$ if $n$ is:
  • ✓
    a positive integer
  • B
    an even positive integer
  • C
    an odd positive integer
  • D
    None of these
Answer
Correct option: A.
a positive integer
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MCQ 1241 Mark
What will be $P(k + 1)$ for $P(n) = n^3(n + 1)?$
  • A
    $ (k+1)^4$
  • ✓
    $ k^4+5 k^3+9 k^2+7 k+2 $
  • C
    $ k^4+6 k^3+9 k^2+7 k+2 $
  • D
    $k^4+3 k^3+9 k^2+6 k+2 $
Answer
Correct option: B.
$ k^4+5 k^3+9 k^2+7 k+2 $
$P(n)=n^3(n+1)$
$P(k+1)=(k+1)^3(k+1+1)$
$P(k+1)=\left(k^3+3 k^2+3 k+1\right)(k+2)$
$P(k+1)=k^4+3 k^3+3 k^2+k+2 k^3+6 k^2+6 k+2$
$P(k+1)=k^4+5 k^3+9 k^2+7 k+2$
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MCQ 1251 Mark
For all $n \in N, 3n^5+ 5n^3+ 7n$ is divisible by:
  • A
    $5$
  • ✓
    $15$
  • C
    $10$
  • D
    $3$
Answer
Correct option: B.
$15$
Given number $= 3n^5+ 5n^2+ 7n$
Let $n = 1, 2, 3, 4, ……..$
$3 n^5+5 n^3+7 n$
$=3 \times 1^2+5 \times 1^3+7 \times 1$
$=3+5+7$
$=15 $
$3 n^5+5 n^3+7 n$
$=3 \times 2^5+5 \times 2^3+7 \times 2$
$=3 \times 32+5 \times 8+7 \times 2$
$=96+40+14$
$=150$
$=15 \times 10 $
$3 n^5+5 n^3+7 n$
$=3 \times 3^5+5 \times 3^3+7 \times 3$
$=3 \times 243+5 \times 27+7 \times 3$
$=729+135+21=885$
$=15 \times 59 $
Since, all these numbers are divisible by $15$ for $n = 1, 2, 3, …..$
So, the given number is divisible by $15$
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MCQ 1261 Mark
Let $P(n)$ be the statement representing the sum of next three successive natural numbers of $n,$ $\forall\text{n}\in\text{N},$ then the smallest value of n to which $P(n)$ is divisible by $9$ is:
  • ✓
    $1$
  • B
    $3!$
  • C
    $3$
  • D
    $9!$
Answer
Correct option: A.
$1$
As $P(n) = (n + 1) + (n + 2) + (n + 3)$
$P(n) = 3n + 3 + 3$
$P(n) = 3(n + 2)$
$\therefore P(1) = 3(3) = 9$
Which is divisible by $9$
$\therefore$ least value of $n$ is $1.$
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MCQ 1271 Mark
For all $\text{n}\in\text{N},3 \times 5^{2n+1}+ 2^{3n+1}$ is divisible by:
  • A
    $19$
  • ✓
    $17$
  • C
    $23$
  • D
    $25$
Answer
Correct option: B.
$17$
$3.5^{2n+1}+ 2^{3n+1}$ is divisible by $17, \text{n}\in\text{N}$
Step $1: 3.5^{2(1)+1}+ 2^{3(1)+1}$
$3.5^3+ 2^4= 391$
Step $2:$ Assuming True for $n = k$
Hence, it is proved that $3.5^{2n+1}+ 2^{3n+1}$ is divisible by $17.$
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MCQ 1281 Mark
The product of three consecutive natural numbers is divisible by:
  • A
    3
  • B
    8
  • C
    6
  • ✓
    (A) and (C) both
Answer
Correct option: D.
(A) and (C) both
d. (A) and (C) both
Solution:
Let n, n + 1, n + 2 be three consecutive natural numbers and P(n) be their product. Then,
P(n) = n(n + 1)(n + 2)
We have,
P(1) = 1 × 2 × 3 = 6, which is divisible by 3 and 6.
P(2) = 2 × 3 × 4 = 24, which is divisible by 3, 8 and 6.
P(3) = 3 × 4 × 5 = 60, which is divisible by 3 and 6.
P(4)= 4 × 5 × 6 = 120, which is divisible by 3, 8 and 6.
Hence, P(n) is divisible by 3 and 6 for all n ∈ N.
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MCQ 1291 Mark
Let $f(n)$ equals to the sum of the cubes of three consecutive natural numbers. $f(n)$ leaves the remainder zero when divided by:
  • A
    $11$
  • ✓
    $9$
  • C
    $99$
  • D
    None of these
Answer
Correct option: B.
$9$
Given that $f(n)=(n-1)^3+n^3+(n+1)^3=3 n^3+6 n$
Put $n=1$, to obtain $f(1)=3.1^3+6.1=9$
Therefore, $f(1)$ is divisible by $9$
Assume that for $n=k, f(k)=3 k^3+6 k$ is divisible by $9$
Now, $f(k+1)=3(k+1)^3+6(k+1)$
$=3 k^3+6 k+9\left(k^2+k+1\right)$
$=f(k)+9\left(k^2+k+1\right)$
Since, $f(k)$ is divisible by $9$
Therefore, $f(k + 1)$ is divisible by $9$
And from the principle of mathematical induction $f(n)$ is divisible by $9$ for all $n \in N.$
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MCQ 1301 Mark
Let $P(n): n^2+ n + 1$ is an even integer. If $P(k)$ is assumed true $\Rightarrow p(k + 1)$ is true. Therefore, $P(n)$ is true:
  • A
    for $n > 1$
  • ✓
    for all $n \in N$
  • C
    for $n > 2$
  • D
    None of these
Answer
Correct option: B.
for all $n \in N$
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MCQ 1311 Mark
If $n \in N,$ then the highest positive integerwhich divides $n(n – 1)(n – 2)$ is:
  • A
    $3$
  • B
    $6$
  • ✓
    $9$
  • D
    $12$
Answer
Correct option: C.
$9$
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MCQ 1321 Mark
Let $P(n): “2n < (1 \times 2 \times 3 \times ... \times n)”.$ Then the smallest positive integer for which $P(n)$ is true is:
  • A
    $1$
  • B
    $2$
  • C
    $3$
  • ✓
    $4$
Answer
Correct option: D.
$4$
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MCQ 1331 Mark
$(n^2+ n)$ is $...........$ for all $n \in N.$
  • ✓
    Even
  • B
    Odd
  • C
    Either even or odd
  • D
    None of these
Answer
Correct option: A.
Even
Concept:
Suppose there is a given statement $P(n)$ involving the natural number $n$ such that
The statement is true for $n = 1,$
i.e., $P(1)$ is true, and
If the statement is true for $n = k ($where $k$ is some positive integer$)$, then the statement is also true for $n = k + 1,$
i.e., truth of $P(k)$ implies the truth of $P(k + 1)$
Then, $P(n)$ is true for all natural numbers $n$
Calculation:
Given:
$P(n)=n^2+n$
Put, $\mathrm{n}=1$
$P(1)=12+1=2 ($Even$)$
Let $P(k)$ is true for $n=k$
$P(k):\left(k^2+k\right)$ is even
$\left(k^2+k\right)=2 m$ for some natural number $m$
Now, $P(k+1)=(k+1)^2+(k+1)=k^2+3 k+2=\left(k^2+k\right)+2(k+1)$
using equation $(1), P(k + 1) = 2m + 2(k + 1) = 2[m + (k + 1)],$ which is even
Hence, $P(k +1)$ is even
$\Rightarrow P(k + 1)$ is true, whenever $P(k)$ is true.
Thus, $P(1)$ is true and $P(k + 1)$ is true, whenever $P(k)$ is true.
Hence, by the principle of Mathematical Induction, $P(n)$ is true for all $n \in N$.
​​​​​​​i.e $p(n) = (n^2+ n)$ is even
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MCQ 1341 Mark
If $10^n+ 3.4^{n+2}+ k$ is divisible by $9$ for all $n \in N,$ then the least positive integral value of $k$ is:
  • ✓
    $5$
  • B
    $3$
  • C
    $7$
  • D
    $1$
Answer
Correct option: A.
$5$
Given that $10^n+ 3.4^{n+2} + k$ is exactly divisible by $9.$
Consider: $P(n) = 10^n+ 3.4^{n+2}+ k$
Substituting $n = 1,$
$P(1) = 10^1+ 3.4^{1+2}+ k$
$= 10 + 3(64) + k$
$= 10 + 192 + k$
$= 202 + k$ is exactly divisible by $9$, the value of $k$ will be $5.$
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MCQ 1351 Mark
If $n$ is an even number, then the digit in the units place of $2^{2n}+ 1$ will be:
  • A
    $5$
  • ✓
    $7$
  • C
    $6$
  • D
    $1$
Answer
Correct option: B.
$7$
Since $2^{2n}$ is even therefore $2^{2n} + 1$ is odd,
therefore digit at unit place should be odd, rejecting option $3.$
Put $n = 2,$ we get $2^{2n}+ 1 = 17$,
Hence digit should be $7$
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MCQ 1361 Mark
Let $T(k)$ be the statement $1 + 3 + 5 +...+ (2k – 1) = k + 10$ Which of the following is correct?
  • A
    $T(1)$ is true
  • ✓
    $T(k)$ is true $\Rightarrow T(k + 1)$ is true
  • C
    $T(n)$ is true for all $\text{n ÎN}$
  • D
    All above are correct
Answer
Correct option: B.
$T(k)$ is true $\Rightarrow T(k + 1)$ is true
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MCQ 1371 Mark
Let $P(n) = 5^n- 2^n. P (n)$ is divisible by $3\lambda$ where $\lambda$ and $n$ both are odd positive integers, then the least value of $n$ and $\lambda$ will be.
  • A
    $13$
  • B
    $11$
  • ✓
    $1$
  • D
    $5$
Answer
Correct option: C.
$1$
$5^n- 2^n$ is divisible by $5 - 2 = 3$ always$...$
Putting $\text{n}=\lambda=1$ which is the least odd positive integer, this works to be true.
Hence Option $C$
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MCQ 1381 Mark
The sum of the series 1 + 2 + 3 + 4 + 5 +………..n is:
  • A
    $\text{n}(\text{n} + 1)$
  • B
    $\frac{(\text{n} + 1)}{2}$
  • C
    $\frac{\text{n}}{2}$
  • ✓
    $\frac{\text{n}(\text{n} + 1)}{2}$
Answer
Correct option: D.
$\frac{\text{n}(\text{n} + 1)}{2}$
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MCQ 1391 Mark
If $n$ is a positive integer, then $2. 42n + 1 + 33n + 1$ is divisible by:
  • A
    $2$
  • B
    $7$
  • ✓
    $11$
  • D
    $27$
Answer
Correct option: C.
$11$
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MCQ 1401 Mark
If $P(n)$ is a statement such that $P(3)$ is true. Assuming $P(k)$ is true $\Rightarrow p(k + 1)$ is true for all $\text{k} \geq 3$, then $P(n)$ is true:
  • A
    for all $n$
  • ✓
    for $n ≥ 3$
  • C
    for $n > 4$
  • D
    None of these
Answer
Correct option: B.
for $n ≥ 3$
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MCQ 1411 Mark
Let $p(n)=x\left(x^{n-1}-n \cdot a^{n-1}+a^n(n-1)\right)$ is divisible by $(x-a)^2$ for:
  • A
    $n > 1$
  • B
    $n > 2$
  • C
    $\forall n \in N$
  • ✓
    None of these
Answer
Correct option: D.
None of these
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MCQ 1421 Mark
The inequality $n ! > 2^{n-1}$ is true for:
  • ✓
    $n > 2$
  • B
    $n \in N$
  • C
    $n > 3$
  • D
    None of these
Answer
Correct option: A.
$n > 2$
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MCQ 1431 Mark
For all $n \in N, 3.5^{2n+1}+ 2^{3n+1}$ is divisible by:
  • A
    $19$
  • ✓
    $17$
  • C
    $23$
  • D
    $25$
Answer
Correct option: B.
$17$
Let $P(n)$ be the statement that $3.5^{2n+1} + 2^{3n+1} $ is divisible by $17$
If $n = 1,$ then given expression $= 3 \times 5^3+ 2^4+ 375 + 16 = 391 = 17 \times 23$, divisible by $17.$
$P(1)$ is true
Assume that $P(k)$ is true.
$3.5^{2k+1}+2^{3k+1}$ is divisible by $17 .$
$ 3.5^{2k}=1+2^{3k+1}=17 {~m}$ where $m \in N$
$ 3.5^{2(k+1)+1}+23^{(k+1)+1}$
$ =3.5^{2k+1} \times 5^2+2^{3k+1} \times 2^3$
$ =25^{(17m-23k+1)}+8.2^{3k+1}​​​​​​​$
$ =425m-25.2^{3k+1}+8.2^{3k+1}​​​​​​​$
$=425m-17.2^{3k+1} $
$ =17\left(25 \mathrm{~m}-2^{3 \mathrm{k}+1}\right)$, divisible by $17$
$P(k + 1)$ is true by Principle of Mathematical Induction
$P(n)$ is true for all $n \in N. 3.5^{2n+1}+ 2^{3n+1}$ is divisible by $17$ for all $n \in N$
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MCQ 1441 Mark
Let $S(K) = 1+ 3 + 5…+ (2K-1) = 3+ K^2$. Then which of the following is true:
  • A
    Principle of mathematical induction can be used to prove the formula
  • ✓
    $S(K) \Rightarrow S(K +1)$
  • C
    $S(K) /\Rightarrow S(K +1)$
  • D
    $S(1)$ is correct
Answer
Correct option: B.
$S(K) \Rightarrow S(K +1)$
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MCQ 1451 Mark
If $p$ is a prime number, then $np−n$ is divisible by $p$ for all $n,$ where:
  • ✓
    $N \in N.$
  • B
    $N$ is odd natural number.
  • C
    $N$ is even natural number.
  • D
    $N$ is not a composite number.
Answer
Correct option: A.
$N \in N.$
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MCQ 1461 Mark
If $1+5+12+22+35+...$ to  $n$ terms $=\frac{\text{n}^2(\text{n+1)}}{2},n^{th}$ term of series is:
  • A
    $\frac{\text{n}(4\text{n}-1)}{3}$
  • ✓
    $\frac{\text{n}(3\text{n}-1)}{2}$
  • C
    $\frac{\text{n}(3\text{n}+1)}{2}$
  • D
    $\frac{\text{n}(4\text{n}+1)}{2}$
Answer
Correct option: B.
$\frac{\text{n}(3\text{n}-1)}{2}$
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MCQ 1481 Mark
$f P(n)$ is a statement $(n \in N)$ such that, if $P(k)$ is true, $P(k + 1)$ is true for $k\in N,$ then $P(n)$ is true:
  • ✓
    for all $n$
  • B
    for all $n > 1$
  • C
    for all $n > 2$
  • D
    Nothing can be said
Answer
Correct option: A.
for all $n$
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MCQ 1491 Mark
The smallest positive integer $n$ for which $n! <\Big(\frac{\text{n+1}}{2}\Big)^\text{n}$ holds, is:
  • A
    $1$
  • ✓
    $2$
  • C
    $3$
  • D
    $4$
Answer
Correct option: B.
$2$
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MCQ 1501 Mark
If $P(n) = 2 + 4 + 6 + .....+ 2n, n Î N ,$ then $P(k) = k(k +1) + 2 \Rightarrow P(k +1) = (k +1)(k + 2) + 2$ for all $k Î N .$ So we can conclude that $P(n) = n(n +1) + 2$ for $D:$
  • A
    $\text{n}\geq1$
  • B
    $\text{n}\geq2$
  • C
    $\text{n}\geq3$
  • ✓
    $\text{none}\text{ of}\text{ these}$
Answer
Correct option: D.
$\text{none}\text{ of}\text{ these}$
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MCQ 1511 Mark
Let $P(n) : n^2+ n + 1$ is an even integer. If $P(k)$ is assumed true $Þ P(k + 1)$ is true. Therefore, $P(n)$ is true:
  • A
    for $n > 1$
  • B
    for all $n \in N$
  • C
    for $n > 2$
  • ✓
    None of these
Answer
Correct option: D.
None of these
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MCQ 1521 Mark
$\{1/(3 ∙ 5)\} + \{1/(5 ∙ 7)\} + \{1/(7 ∙ 9)\} + ……. + 1/\{(2n + 1) (2n + 3)\}:$
  • A
    $n/(2n + 3)$
  • B
    $n/\{2(2n + 3)\}$
  • ✓
    $n/\{3(2n + 3)\}$
  • D
    $n/\{4(2n + 3)\}$
Answer
Correct option: C.
$n/\{3(2n + 3)\}$
Let the given statement be $P(n).$ Then,
$P(n): \{1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9)\} + ……. + 1/\{(2n + 1)(2n + 3)\} = n/\{3(2n + 3)\}.$
Putting $n = 1$ in the given statement, we get
and $\text{LHS} = 1/(3 ∙ 5) = 1/15$ and $\text{RHS} = 1/\{3(2 \times 1 + 3)\} = 1/15.$
$\text{LHS = RHS}$
Thus, $P(1)$ is true.
Let $P(k)$ be true. Then,
$P(k): \{1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + …….. + 1/\{(2k + 1)(2k + 3)\} = k/\{3(2k + 3)\} ….. (i)$
Now, $1/(3 ∙ 5) + 1/(5 ∙ 7) + ..…… + 1/[(2k + 1)(2k + 3)] + 1/[\{2(k + 1) + 1\}2(k + 1) + 3$
$= \{1/(3 ∙ 5) + 1/(5 ∙ 7) + ……. + [1/(2k + 1)(2k + 3)]\} + 1/\{(2k + 3)(2k + 5)\}$
$= k/[3(2k + 3)] + 1/[2k + 3)(2k + 5)] [$using $(i)]$
$= \{k(2k + 5) + 3\}/\{3(2k + 3)(2k + 5)\}$
$= (2k^2 + 5k + 3)/[3(2k + 3)(2k + 5)]$
$= \{(k + 1)(2k + 3)\}/\{3(2k + 3)(2k + 5)\}$
$= (k + 1)/\{3(2k + 5)\}$
$= (k + 1)/[3\{2(k + 1) + 3\}]$
$= P(k + 1): 1/(3 ∙ 5) + 1/(5 ∙ 7) + …….. + 1/[2k + 1)(2k + 3)] + 1/[\{2(k + 1) + 1\}\{2(k + 1) + 3\}]$
$= (k + 1)/\{3\{2(k + 1)\} + 3\}]$
$\Rightarrow P(k + 1)$ is true, whenever $P(k)$ is true.
Thus, $P(1)$ is true and $P(k + 1)$ is true, whenever $P(k)$ is true.
Hence, by the principle of mathematical induction, $P(n)$ is true for $n \in N$
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MCQ 1531 Mark
Choose the correct answer. If $10^n+ 3 \times 4^{n+2} + k$ is divisible by $9$ for all $n \in N,$ then the least positive integral value of $k$ is:
  • ✓
    $5$
  • B
    $3$
  • C
    $7$
  • D
    $1$
Answer
Correct option: A.
$5$
Let $P(n) = 10^n+ 3 \times 4^{n+2}+ k$ is divisible by $9, \forall\text{ n}\in\text{N}$
$P(1) = 10^1+ 3 \times 4^{1+2}+ k$
$= 10 + 3 \times 64 + k$
$= 10 + 192 + k$
$= 202 + k$ must be divisible by $9.$
If $(202 + k)$ is divisible by $9$ then $k$ must be equal to $5.$
$202 + 5 = 207$ which is divisible by $9.$
$=\frac{207}{9}$
$=23$
So, the least positive integral value of $k = 5$
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MCQ 1541 Mark
For each $n \in N, 10^{2n-1}+ 1$ is divisible by
  • ✓
    $11$
  • B
    $13$
  • C
    $9$
  • D
    None of these
Answer
Correct option: A.
$11$
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MCQ 1551 Mark
For any natural number $n, 2^{2n} - 1$ is divisible by:
  • A
    $2$
  • ✓
    $3$
  • C
    $4$
  • D
    $5$
Answer
Correct option: B.
$3$
Let $P(n)=2^{2 n}-1$
Substituting $n = 1, 2, 3,….$
$P(1)=2^{2(1)}-1=4-1=3$
This is divisible by $3.$
$P(2)=2^{2(2)}-1=16-1=15$
This is divisible by $3.$
$P(3)=2^{2(3)}-1=256-1=255$
This is also divisible by $3.$
Assume that $P(n)$ is true for some natural number $k,$
i.e., $P(k): 2^{2k}- 1$ is divisible by $3,$
i.e., $2^{2k}- 1 = 3q$,
where $q \in N$
Now,
$P(k+1): 2^{2(k+1)}-1$
$=2^{2 k+2}-1$
$=2^{2 k} \times 2^2-1$
$=2^{2 k} \times 4-1$
$=3.2^{2 k}+\left(2^{2 k}-1\right)$
$=3.2^{2 k}+3 q$
$=3\left(2^{2 k}+q\right)=3m,$
where $m \in N$
Thus $P(k + 1)$ is true, whenever $P(k)$ is true.
Therefore, for any natural number $n, 2^{2n-1}$ is divisible by $3$
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MCQ 1571 Mark
Statement$-l:$ For every natural number $\text{n}\geq2,\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{\text{n}}}>\sqrt{\text{n}}$
Statement$-2:$ For every natural number $\text{n}\geq2,\sqrt{\text{n}(\text{n}+1)}>\text{n}+1.$
  • A
    Statement$-1$ is true, Statement$-2$ is true; Statement$-2$ is a correct explanation for Statement$-1.$
  • B
    Statement$-1$ is true, Statement$-2$ is false.
  • C
    Statement$-1$ is false, Statement$-2$ is true.
  • ✓
    Statement$-1$ is true, Statement$-2$ is true; Statement$-2$ is not a correct explanation for Statement$-1$.
Answer
Correct option: D.
Statement$-1$ is true, Statement$-2$ is true; Statement$-2$ is not a correct explanation for Statement$-1$.
$\text{P}\ (\text{n})\ =\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{\text{n}}}$
$\text{P}\ (2)\ =\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}>\sqrt{2}$
Let us assume that $P(k)$
$=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{\text{k}}}>\sqrt{\text{k}}$ is true
$\therefore\ \text{P (k + 1)}=\ \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{\text{k}}}+\frac{1}{\sqrt{\text{k + 1}}}>\sqrt{\text{k+1}}$ has to be true
$\text{L.H.S}>\sqrt{\text{k}}+\frac{1}{\sqrt{\text{k + 1}}}=\frac{\sqrt{\text{k}\ (\text{k + 1)}+1}}{\sqrt{\ \text{k + 1}}}$
since $\sqrt{\text{k}\ (\text{k} + 1)}>\text{k}\ (\forall\text{k}\geq0)$
$\therefore\frac{\sqrt{\text{k}\ (\text{k + 1)}+1}}{\sqrt{\ \text{k + 1}}}>\frac{\text{k + 1}}{\sqrt{\text{k + 1}}}=\sqrt{\text{k + 1}}$
Let $\text{P}\ (\text{n})=\sqrt{\text{n}\ (\text{n + 1})}<(\text{k + 1})$
Statement$-1$ is correct.
$\text{P}\ (2)=\sqrt{2\times3}<3$
If $\text{P}\ \text{(k)}=\sqrt{\text{k (k + 1)}}<\text{(k + 1)}$ is true
Now $\text{P}\ \text{(k + 1)}=\sqrt{\text{(k + 1)}(\text{k + 2})}<\text{(k + 2)}$ has to be true
since $\text{(k + 1)}<\text{k + 2}$
$\therefore\sqrt{\text{(k + 1)}\text{(k + 2)}}<\text{(k + 2)}$
Hence Statement$-2$ is not correct explanation of Statement$-1.$
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MCQ 1581 Mark
For all $n \in N, 3n^5+ 5n^3+ 7n$ is divisible by:
  • A
    $5$
  • ✓
    $15$
  • C
    $10$
  • D
    $3$
Answer
Correct option: B.
$15$
Given number $=3 n^5+5 n^2+7 n $
Let $ n=1,2,3,4,...... $
$ 3 n^5+5 n^3+7 n$
$=3 \times 1^2+5 \times 1^3+7 \times 1$
$=3+5+7$
$=15 $
$ 3 n^5+5 n^3+7 n$
$=3 \times 2^5+5 \times 2^3+7 \times 2$
$=3 \times 32+5 \times 8+7 \times 2$
$=96+40+14$
$=15 \times 10$
$=150$
$ 3 n^5+5 n^3+7 n$
$=3 \times 3^5+5 \times 3^3+7 \times 3$
$=3 \times 243+5 \times 27+7 \times 3$
$=729+135+21$
$=15 \times 59$
$=885$
Since, all these numbers are divisible by $15$ for $n = 1, 2, 3, …..$
So, the given number is divisible by $15.$
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MCQ 1621 Mark
If $p(n): 49^\text{n}+16^{\text{n}}\lambda$ is divisible by $64$ for $\text{n}\in\text{N}$ is true, then the least negative integral value of $\lambda$ is:
  • A
    $-3$
  • B
    $-2$
  • ✓
    $-1$
  • D
    $-4$
Answer
Correct option: C.
$-1$
$ (49)^n+16 n-1 $
$ \Rightarrow(1+48)^n+16 n-1 $
$ \Rightarrow 1+48 n+\ldots 48^n+16 n-1 $
$ \Rightarrow 64 n+n C_2(48)^2+n C_3(48)^3+\ldots+(48)^n $
$ \Rightarrow 64\left(n+n C_2(6)^2+n C_3(6)^3 48+\ldots+(6)^n 8^{n-2}\right)$
$\therefore 49^n+ 16n - 1$ is divisible by $64$
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MCQ 1631 Mark
$7^{2n}+ 3^{n-1}⋅ 2^{3n-3}$ is divisible by:
  • A
    $24$
  • ✓
    $25$
  • C
    $9$
  • D
    $13$
Answer
Correct option: B.
$25$
Let $P(1) =7^{2n}+ 3^{n-1}⋅ 2^{3n-3}$
$P(1) = 50 \Rightarrow$ Divisible by $25$
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MCQ 1641 Mark
The sum of the series $1^2 + 2^2 + 3^2 + ……….. n^2$ is:
  • A
    $\text{n}(\text{n+1}(2\text{n+1)}$
  • B
    $\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{2}$
  • C
    $\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{3}$
  • ✓
    $\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{6}$
Answer
Correct option: D.
$\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{6}$
Given, series is $1^2 + 2^2 + 3^2 + ……….. n^2$
$\text{sum=}\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{6}$
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MCQ 1651 Mark
Let $f(n) = 8n - 3n,$ if $n$ is odd natural number then $f(n)$ is divisible by:
  • A
    $2$
  • B
    $3$
  • ✓
    $5$
  • D
    None of these.
Answer
Correct option: C.
$5$
$f(n) = 8^n- 3^n$
Since, $n$ is odd
for $n = 1,$ we get $f(1) = 8^1− 3^1= 5$
for $n = 3,$ we get $f(3) = 8^3- 3^3= 5(97)$
for $n = 5,$ we get $f(3) = 8^5- 3^5= 5(6505)$
Therefore, by induction we can say that $f(n)$ is divisible by $5$ for odd $n.$
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MCQ 1661 Mark
In the following question, assuming the given statements to be true, find which of the conclusion among the given conclusions is/are definitely true and then give your answer accordingly.
Statement: $A ≥ P > T; V < B ≥ X; P = S; B = T$
Conclusion: $I. A > XII. P < B:$
  • A
    None is True.
  • B
    Both $I$ and $II$ are True.
  • C
    Only $II$ is True.
  • ✓
    Only $I$ is True.
Answer
Correct option: D.
Only $I$ is True.
Given statement: $A ≥ P > T; V < B ≥ X; P = S; B = T$
On combining: $A ≥ P = S > T = B ≥ X; V < B$
Conclusions:
$I. A > X \rightarrow$ True $(A \geq P = S > T = B \geq X)$
$II. P < B \rightarrow$ False $(P > T = B)$
Hence, only $I$ is True.
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MCQ 1671 Mark
The value of $(1+3/1)(1+5/4)(1+7/9)…(1+2n+1/n^2)$ is:
  • A
    $(n + 1)$
  • B
    $(n + 1)^2$
  • C
    $42(n+1)^2$
  • ✓
    None of these
Answer
Correct option: D.
None of these
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MCQ 1681 Mark
If $P(n)$ is statement such that $P(3)$ is true. assuming $P(k)$ is true $\Rightarrow P(k + 1)$ is true for all $k \geq 3,$ then $P(n)$ is true:
  • A
    for all $n$
  • ✓
    for $n ≥ 3$
  • C
    for $n ≥ 4$
  • D
    none of these
Answer
Correct option: B.
for $n ≥ 3$
Concepts:
Suppose there is a given statement $P(n)$ involving the natural number $n$ such that
The statement is true for $n = 1,$
i.e., $P(1)$ is true, and
If the statement is true for $n = k ($where $k$ is some positive integer$)$, then the statement is also true for $n = k + 1,$ i.e., the truth of $P(k)$ implies the truth of $P(k + 1).$
Then, $P(n)$ is true for all natural numbers $n$
In the given question $P(n)$ is true for $n = 3$
Assuming $P(k)$ is true
$\Rightarrow P(k + 1)$ is true for $k \geq 3$
Hence, By the Principle of Mathematical induction $P(n)$ is true for all $n \geq 3.$
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MCQ 1701 Mark
$1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + …….. + 1/{n(n + 1)(n + 2)}:$
  • ✓
    $\{n(n + 3)\}/\{4(n + 1)(n + 2)\}$
  • B
    $(n + 3)/\{4(n + 1)(n + 2)\}$
  • C
    $n/\{4(n + 1)(n + 2)\}$
  • D
    None of these
Answer
Correct option: A.
$\{n(n + 3)\}/\{4(n + 1)(n + 2)\}$
Let $P (n): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……. + 1/\{n(n + 1)(n + 2)\} = \{n(n + 3)\}/\{4(n + 1)(n + 2)\}$
Putting $n = 1$ in the given statement, we get
$\text{LHS} = 1/(1 ∙ 2 ∙ 3) = 1/6$ and $\text{RHS} = \{1 \times (1 + 3)\}/[4 \times (1 + 1)(1 + 2)] = ( 1 \times 4)/(4 \times 2 \times 3) = 1/6.$
Therefore $\text{LHS = RHS.}$
Thus, the given statement is true for $n = 1,$
i.e., $P(1)$ is true.
Let $P(k)$ be true. Then,
$P(k): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + … + 1/\{k(k + 1) (k + 2)\} = \{k(k + 3)\}/\{4(k + 1) (k + 2)\}.…(i)$
Now, $1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………….. + 1/\{k(k + 1) (k + 2)\} + 1/\{(k + 1) (k + 2) (k + 3)\}$
$= [1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………..…. + 1/{ k(k + 1) (k + 2}] + 1/{(k + 1)(k + 2) (k + 3)}$
$= [\{k(k + 3)\}/\{4(k + 1)(k + 2)\} + 1/\{(k + 1)(k + 2)(k + 3)\}] [$using $(i)]$
$= {k(k + 3)^2 + 4}/\{4(k + 1)(k + 2) (k + 3)\}$
$= (k^3 + 6k^2 + 9k + 4)/\{4(k + 1) (k + 2) (k + 3)\}$
$= {(k + 1) (k + 1) (k + 4)}/\{4 (k + 1) (k + 2) (k + 3)\}$
$= \{(k + 1) (k + 4)\}/\{4(k + 2) (k + 3)\}$
$\Rightarrow P(k + 1): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……….….. + 1/\{(k + 1) (k + 2) (k + 3)\}$
$= \{(k + 1) (k + 2)\}/\{4(k + 2) (k + 3)\}$
$\Rightarrow P(k + 1)$ is true, whenever $P(k)$ is true.
Thus, $P(1)$ is true and $P(k + 1)$ is true, whenever $P(k)$ is true.
Hence, by the principle of mathematical induction, $P(n)$ is true for all $n \in N.$
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MCQ 1711 Mark
If $x^{n-1}$ is divisible by $x - k,$ then the least positive integral value of $k$ is:
  • ✓
    $1$
  • B
    $2$
  • C
    $3$
  • D
    $4$
Answer
Correct option: A.
$1$
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MCQ 1721 Mark
$(1^2 + 2^2 + …… + n^2) ...........$ for all values of $n \in N:$
  • A
    $= n^3/3$
  • B
    $< n^3 /3$
  • ✓
    $> n^3/3$
  • D
    None of these
Answer
Correct option: C.
$> n^3/3$
Let $P(n): (1^2 + 2^2 + ….. + n^2) > n^3/3.$
When $= 1, \text{LHS} = 1^2 = 1$ and $\text{RHS} = 1^3/3 = 1/3.$
Since $1 > 1/3,$ it follows that $P(1)$ is true.
Let $P(k)$ be true. Then,
$P(k): (1^2 + 2^2+ ….. + k^2 ) > k^3/3 …. (i)$
Now,
$1^2 + 2^2 + ….. + k^2+ (k + 1)^2$
$= \{1^2 + 2^2 + ….. + k^2 + (k + 1)^2$
$> k^3/3 + (k + 1)^3 [$using $(i)]$
$= 1/3 ∙ (k^3 + 3 + (k + 1)^2) = 1/3 ∙ \{k^2 + 3k^2 + 6k + 3\}$
$= 1/3[k^3 + 1 + 3k(k + 1) + (3k + 2)]$
$= 1/3 ∙ [(k + 1)^3 + (3k + 2)]$
$> 1/3(k + 1)^3P(k + 1):$
$1^2 + 2^2 + ….. + k^2 + (k + 1)^2$
$> 1/3 ∙ (k + 1)^3$
$P(k + 1)$ is true, whenever $P(k)$ is true.
Thus $P(1)$ is true and $P(k + 1)$ is true whenever $p(k)$ is true.
Hence, by the principle of mathematical induction, $P(n)$ is true for all $n \in N.$
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MCQ 1731 Mark
For every positive integer $n, 7^n– 3^n$ is divisible by:
  • A
    $3$
  • ✓
    $4$
  • C
    $7$
  • D
    $5$
Answer
Correct option: B.
$4$
Let $P(n) = 7^n- 3^n$
Substituting $n = 1, 2, 3,…$
$P(1) = 7^1– 3^1= 7 - 3 = 4 $
$P(2) = 7^2- 3^2= 49 - 9 = 40 $
$P(3) = 7^3-3^3= 343 - 27 = 316 $
Thus, for every positive integer $n, 7^n-3^n $ is divisible by $4.$
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MCQ 1741 Mark
For all $n \in N, 3 \times 5^{2n+1}+ 2^{3n+1}$ is divisible by:
  • A
    $19$
  • ✓
    $17$
  • C
    $23$
  • D
    $25$
Answer
Correct option: B.
$17$
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MCQ 1751 Mark
What is the sum of $1 + 2 + 3 + ... n\ ?$
  • A
    $\frac{\text{n+1}}{2}$
  • B
    $\frac{\text{n}}{2}$
  • ✓
    $\frac{\text{n}(\text{n+1)}}{2}$
  • D
    $\frac{\text{n}(\text{n+2)}}{2}$
Answer
Correct option: C.
$\frac{\text{n}(\text{n+1)}}{2}$
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MCQ 1761 Mark
If $49^n+ 16^n+ k$ is divisible by $64$ for $n \in N,$ then the least negative integral value of $k$ is:
  • ✓
    $-1$
  • B
    $-2$
  • C
    $-3$
  • D
    $-4$
Answer
Correct option: A.
$-1$
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MCQ 1781 Mark
If $n(n^2 − 1)$ is divisible by $24,$ then which of the following statements is true?
  • ✓
    $n$ can be any odd integral value.
  • B
    $n$ can be any integral value.
  • C
    $n$ can be any even integral value.
  • D
    $n$ can be any rational number.
Answer
Correct option: A.
$n$ can be any odd integral value.
$n(n^2 − 1) = n ∗ (n − 1) ∗ (n + 1)$
If $n$ is even, then $n − 1$ and $n + 1$ will be odd,
therefore $n(n^2 − 1)$ is not divisible by $4$ and therefore not divisible by $24.$
Hence $n$ has to be an odd integer.
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MCQ 1791 Mark
If $P(n) = 2 + 4 + 6 +…+ 2n, n \in N ,P(k) = k(k + 1) + 2 \Rightarrow P(k + 1) = (k + 1)(k + 2) + 2$ for all $k \in N$. So, we can conclude that $P(n) = n (n + 1) 2$ for:
  • A
    all $n \in N$
  • B
    $n > 1$
  • C
    $n > 2$
  • ✓
    nothing can be said
Answer
Correct option: D.
nothing can be said
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MCQ 1801 Mark
For any natural number $n, 7^n– 2^n$ is divisible by
  • A
    $3$
  • B
    $4$
  • ✓
    $5$
  • D
    $7$
Answer
Correct option: C.
$5$
Given, $7^n– 2^n$
Let $n = 1$
$7^n-2^n$
$=7^1-2^1$
$=7-2$
$=5$
which is divisible by $5$
Let $n = 2$
$7^n-2^n$
$=7^2-2^2$
$=49-4$
$=45$
which is divisible by $5$
Let $n = 3$
$7^n-2^n$
$=7^3-2^3$
$=343-8$
$=335$
which is divisible by $5$
Hence, for any natural number $n, 7^n-2^n$ is divisible by $5$
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MCQ 1811 Mark
For all positive integers $n > 1, \{x(x^{n-1}- na^{n-1} \})$ $a^n(n-1)$ is divisible by:
  • A
    $(x - a)^2$
  • ✓
    $x - a$
  • C
    $2(x - a)$
  • D
    $x + a$
Answer
Correct option: B.
$x - a$
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MCQ 1821 Mark
If $P(n) = 2 + 4 + 6 +…. + 2 n, n \in N,$ then $P(k) = k(k + 1) + 2 \Rightarrow P(k + 1) = (k + 1) (k + 2) + 2$ for all $k \in N.$ So, we can conclude that $P(n) = n(n + 1) + 2$ for:
  • A
    all $n \in N$
  • B
    $n > 1$
  • C
    $n > 2$
  • ✓
    nothing can be said
Answer
Correct option: D.
nothing can be said
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MCQ 1831 Mark
If $x^n- 1$ is divisible by $x - k,$ then the least positive integral value of $k$ is:
  • ✓
    $1$
  • B
    $2$
  • C
    $3$
  • D
    $4$
Answer
Correct option: A.
$1$
Given,
$P(n): x^n- 1$ is divisible by $x - k$
Let us substitute $n = 1, 2, 3,..$
$ \Rightarrow P(1): x-1 $
$ \Rightarrow P(2): x^2-1=(x-1)(x+1) $
$ \Rightarrow P(3): x^3-1=(x-1)\left(x^2+x+1\right) $
$ \Rightarrow P(4): x^4-1=\left(x^2-1\right)\left(x^2+1\right)=(x-1)(x+1)\left(x^2+1\right) $
Therefore, the least positive integral value of $k$ is $1.$
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MCQ 1841 Mark
Let $P(n) : n^2 + n + 1$ is an even integer. If $P(k)$ is assumed true then $P(k + 1)$ is true. Therefore $P(n)$ is true.
  • A
    for $n > 1$
  • B
    for all $n Î N$
  • C
    for $n > 2$
  • ✓
    none of thes
Answer
Correct option: D.
none of thes
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MCQ 1851 Mark
For all $n \in N, 41^n– 14^n$ is a multiple of:
  • A
    $26$
  • ✓
    $27$
  • C
    $25$
  • D
    None of these
Answer
Correct option: B.
$27$
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