MCQ 11 Mark
If $x^{2 n-1}+y^{2 n-1}$ is divisible by $x + y,$ if $n$ is:
View full question & answer→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
$n + 2$
AnswerCorrect 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 $n^2+ n + 1$
View full question & answer→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 $n?$
AnswerCorrect option: A. For all $n$
View full question & answer→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
$n > 2$
- ✓
AnswerConcepts:
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 $P(n) = n(n + 1) + 2$
View full question & answer→MCQ 51 Mark
For natural number $n, 2^n, (n – 1) ! < ,$ if:
- A
$n < 2$
- ✓
$n > 2$
- C
$n ≥ 2$
- D
AnswerCorrect option: B. $n > 2$
View full question & answer→MCQ 61 Mark
Let $P(n) = 2^{3n}− 7n − 1$ then $P(n)$ is divisible by:
Answer$P(n) = 2^{3n}− 7n − 1 = −1 − 7n + (1 + 7)^n$
$\Rightarrow \ce{P(n)=-1-7 n + \left(1 + n C_1 7 + n C_2 7^2 + ... + n C_n 7^n\right)=n C_2 7^2 + ... + n C_n 7^n}$
$\Rightarrow \ce{P(n)=7^2\left({nC}_2 + {nC}_3 7 + ... + {nC}_{n} 7^{n}-2\right)}$
Therefore, $P(n)$ is divisible by $49.$
View full question & answer→MCQ 71 Mark
For all positive integers $n$, the number $n(n^2 − 1)$ is divisible by:
AnswerGiven,
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 $6$
View full question & answer→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
AnswerCorrect 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.
View full question & answer→MCQ 91 Mark
For all $n \in N, 2.4^{2n+1}+ 3^{3n+1}$ is divisible by:
View full question & answer→MCQ 101 Mark
If $10^{3n}+ 2^{4k+1}× 9 + k,$ is divisible by $11,$ then what is the least positive value of $k?$
Answer$P(n) = 10^{3n} + 2^{4k+1}\times 9 + k$
$P(1) = 10^3+ 2^5\times 9 + k$
$P(1) = 1000 + 288 + k$
$P(1) = 1288 + k$
When $1288$ is divided by $11$, the remainder is $1.$
Therefore, $1287$ is divisible by $11.$
The next number that is divisible is $1298.$
$k = 1298 - 1288$
$k = 10$
View full question & answer→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:
AnswerThe smallest positive integer for which $P(n)$ is $4.$
$P(4) = 2^4 < (1 \times 2 \times 3 \times ... \times 4)$
$P(4) = 16 < 24.$
View full question & answer→MCQ 121 Mark
For all $n \in N, n (n + 1)(n + 5)$ is a multiple of
View full question & answer→MCQ 131 Mark
By principle of mathematical induction, $2^{4n-1}$ is divisible by which of the following?
Answer$P(n)=2^{4 n-1} P(1)=2^3=8$
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 $16.$
View full question & answer→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
AnswerCorrect option: C. all $n > 3$
View full question & answer→MCQ 151 Mark
For each $n \ N \in , 10^{2n-1}+ 1$ is divisible by
View full question & answer→MCQ 161 Mark
If $S_n$ is divisible for every $n,$ then $S_n$ is:
View full question & answer→MCQ 171 Mark
Let $S(k) = 1 + 3 + 5 + (2k – 1) = 3 + K^2$. Then, which of the following is true?
- A
- ✓
- C
- D
Principle of mathematical induction can be used to prove the formula
View full question & answer→MCQ 181 Mark
The inequality $n \ !>2^{n-1}$ is true for
- ✓
$n > 2$
- B
$n\in N$
- C
$n > 3$
- D
AnswerCorrect option: A. $n > 2$
View full question & answer→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:
View full question & answer→MCQ 201 Mark
$n^3+ 5n$ is divisible by which of the following?
Answer$P(n) = n^3 + 5n$
$P(1) = 1 + 5$
$P(1) = 6$
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 $3$
View full question & answer→MCQ 211 Mark
$n(n + 1) (n + 5)$ is a multiple of $........$ for all $n \in N$
AnswerLet $P(n): n(n + 1)(n + 5)$ 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 $n \in N$
View full question & answer→MCQ 221 Mark
If $\forall\text{m}\in\text{N},$ then $11^{m+2}+ 12^{2m-1}$ is divisible by:
AnswerTo 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 $C.$
View full question & answer→MCQ 231 Mark
Let $P (n)$ be the statement $3n > nn.$ If $P (n)$ is true, then $P (n + 1)$ is
AnswerConcepts:
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.
View full question & answer→MCQ 241 Mark
For any natural number $n, 7^n – 2^n$ is divisible by
AnswerGiven, $ 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 $5$
View full question & answer→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
$\ce{mkn = mnk}$
AnswerCorrect 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.
View full question & answer→MCQ 261 Mark
For all $n \ \in N, 2.4^{2n+1}+ 3^{3n+1}$ is divisible by
View full question & answer→MCQ 271 Mark
For each $n \in N, 3^{2n}-1$ is divisible by:
View full question & answer→MCQ 281 Mark
$10n + 3(4^{n+2})+ 5$ is divisible by $(n \in N):$
View full question & answer→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
$\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{3}$
AnswerCorrect option: A. $\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{2}$
View full question & answer→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:
View full question & answer→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
$\text{none}\text{ of}\text{ these}$
AnswerCorrect 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 $n.$
View full question & answer→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
AnswerCorrect option: A. $a(100) ≤ 100$
View full question & answer→MCQ 331 Mark
For each $n \in N,$ the correct statement is
- A
$2n < n$
- B
$nn2 > 2$
- ✓
$n4n < 10$
- D
$2713n > n +$
AnswerCorrect option: C. $n4n < 10$
View full question & answer→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:
Answer$5^n− 2^n$ is divisible by $5 − 2 = 3$ always$...$
Putting $n = \lambda = 1$ which is the least odd positive integer,
this works to be true.
View full question & answer→MCQ 351 Mark
$(2 ∙ 7^N+ 3 ∙ 5^N– 5)$ is divisible by $………..$ for all $N \in N:$
AnswerLet $P(n): (2 ∙ 7^n+ 3 ∙ 5^n– 5)$ 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 $n \in N.$
View full question & answer→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:
Answer$P(1) : 2^1 < 1$
$2< 1$ 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
View full question & answer→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
AnswerCorrect option: C. all $n \in N$
View full question & answer→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.
AnswerCorrect option: A. Both $P(3)$ and $P(5)$ are true.
View full question & answer→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
$-\pi$
View full question & answer→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:
View full question & answer→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:
View full question & answer→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:
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
$P(50)$
- ✓
AnswerGiven, $\ce{P(n) = n(n + 1)}$
$\therefore P(3) = 3.4 = 12($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.
View full question & answer→MCQ 441 Mark
For all $n \in N, 41^n- 14^n$ is a multiple of:
View full question & answer→MCQ 451 Mark
For principle of mathematical induction to be true, what type of number should $'n\ '$ be?
AnswerAccording 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.
View full question & answer→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
$\Big(\frac{\text{n}(\text{n-1)}}{2}\Big)^2$
AnswerCorrect option: B. $\Big(\frac{\text{n}(\text{n+1)}}{2}\Big)^2$
View full question & answer→MCQ 471 Mark
If $P(k) = k^2(k + 3) (k^2– 1)$ is true, then what is $P(k + 1)\ ?$
AnswerCorrect 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 $k.$
$P(k + 1) = (k + 1)^2(k + 1 + 3) ((k + 1)^2– 1)$
$P(k + 1) = (k + 1)^2(k + 4) (k^2+ 1 + 2k – 1)$
$P(k + 1) = (k + 1)^2(k + 4) (k^2+ 2k)$
$P(k + 1) = (k + 1)^2(k + 4) k (k +2)$
View full question & answer→MCQ 481 Mark
For all $n \in N, 3n^5+ 5n^3+ 7n$ is divisible by:
AnswerGiven 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 $1$
View full question & answer→MCQ 491 Mark
If $n \in N,$ then $n(n^2− 1)$ is divisible by:
Answer$\ce{n(n^2 − 1)= n(n − 1)(n + 1)}$
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 $6.$
View full question & answer→MCQ 501 Mark
For every positive integer $n, n7/7 + n5/5 + 2n^3/3 - n/105$ is:
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