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 $n$th terms of the series $3+7+13+21+$………. is
- A
$4 n-1$
- B
$2 n+1$
- ✓
$n^2+n+1$
- D
$n+2$
AnswerCorrect option: C. $n^2+n+1$
- $n^2+n+1$
Solution:
Concept:
Let $S=3+7+13+21+\ldots \ldots \ldots . . . a_{n-1}+a_n \ldots(1)$
and $S=3+7+13+21+\ldots . . . . . a_{n-1}+a_n \ldots(2)$
Substracting Equation (1) and equation (2) we get,
$S-S=3+\left(7+13+21+\ldots \ldots \ldots . . a_{n-1}+a_n\right)-\left(3+7+13+21+\ldots \ldots \ldots . . . . a_{n-1}+a_n\right)$
$\Rightarrow 0=3+(7-3)+(13-7)+(21-13)+\ldots \ldots . . .+\left(a_n-a_{n-1}\right)-a_n$
$\Rightarrow 0=3+\{4+6+8+\ldots \ldots(n-1) \text { terms }\}-a_n$
$\Rightarrow a_n=3+\{4+6+8+\ldots \ldots . .(n-1) \text { terms }\}$
$4+6+8+\ldots . .(n-1)$ terms is an Arithmatic Progression with first term $=4$, common difference $=2$ and no. of terms
$=n-1$
$\mathrm{a}_{\mathrm{n}}=3+\frac{\mathrm{n}-1}{2} \times(2 \times 4+(\mathrm{n}-1-1) \times 2)$
$\mathrm{a}_{\mathrm{n}}=3+\frac{\mathrm{n}-1}{2} \times(8+(\mathrm{n}-2) \times 2)$
$\Rightarrow \mathrm{a}_{\mathrm{n}}=3(\mathrm{n}-1) \times(4+\mathrm{n}-2)$
$\Rightarrow \mathrm{a}_{\mathrm{n}}=3(\mathrm{n}-1) \times(\mathrm{n}-2)$
$\Rightarrow \mathrm{a}_{\mathrm{n}}=\mathrm{n}^2+\mathrm{n}+1$
So, the $\mathrm{n}_{\text {th }}$ term is $\mathrm{n}^2+\mathrm{n}+1$ View full question & answer→MCQ 31 Mark
Let P(n) be a statement and P(n)=P(n+1)∀n ∈ N, then P(n) is true for what values of n?
- ✓
- B
- C
For all n>m , m being a fixed positive integer
- D
View full question & answer→MCQ 41 Mark
If P(n) = 2 + 4 + ......+ 2n, n ϵ N, then P(k) = k(k + 1) + 2 ⇒ P(k) = k(k + 1) + 2 for all k ϵ N. S we can conclude that P(n) = n(n + 1) + 2 for
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:
View full question & answer→MCQ 61 Mark
Let $P(n)=2^{3 n}-7 n-1$ then $P(n)$ is divisible by:
Answer
- 49
Solution:
$P(n)=2^{3 n}-7 n-1=-1-7 n+(1+7)^n$
$\Rightarrow P(n)=-1-7 n+\left(1+n C_1 7+n C_2 7^2+\ldots+n C_n 7^n\right)=n C_2 7^2+\ldots+n C_n 7^n$
$\Rightarrow P(n)=7^2\left(n C_2+n C_3 7+\ldots+n C_n 7^{n-2}\right)$
Therefore, $P(n)$ is divisible by 49 . View full question & answer→MCQ 71 Mark
For all positive integers n, the number n(n² − 1) is divisible by:
AnswerGiven,
number = n(n² - 1)
Let n = 1, 2, 3, 4….
n(n² - 1) = 1(1 - 1) = 0
n(n² - 1) = 2(4 - 1) = 2 × 3 = 6
n(n² - 1) = 3(9 - 1) = 3 × 8 = 24
n(n² - 1) = 4(16 - 1) = 4 × 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 $\mathrm{m}^2>100$, the statement $P(k+1)$ will be true if:
Answer
- P(k) is true
Solution:
$\Rightarrow r^2>100 \Rightarrow r^2+2 r+1>100+2 r+1 \\
& \Rightarrow(r+1)^2>100 \\
& \Rightarrow P(r+1) \text { is true as } \\
& \left.r^2+(2 r+1)>r^2>100 \Rightarrow P(k+1) \text { is true (say } r=k\right)$
$P(k+1)$ is true when every $p(k)$ is
So, In order to prove that $P(k+1)$ is true.
It is sufficient to consider $\mathrm{P}(\mathrm{k})$ is true. View full question & answer→MCQ 91 Mark
For all $n \in N, 2.4^{2 n+1}+3^{3 n+1}$ is divisible by:
View full question & answer→MCQ 101 Mark
If $10^{3 n}+2^{4 k+1} \times 9+k$, is divisible by 11, then what is the least positive value of k?
Answer
- 10
Solution:
$P(n)=10^{3 n}+2^{4 k+1} \times 9+k$
$P(1)=10^3+2^5 \times 9+k$
$P(1)=1000+288+k$
$P(1)=1288+k$
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 $p(n): 2 n<(1 \times 2 \times 3 \times \ldots \times n)$. Then the smallest positive integer for which $p(n)$ is true is:
Answer
- 4
Solution:
The smallest positive integer for which $\mathrm{P}(\mathrm{n})$ is 4 .
$P(4)=2^4<(1 \times 2 \times 3 \times \ldots \times 4)$
$P(4)=16<24$ View full question & answer→MCQ 121 Mark
For all n ∈ N, n (n + 1)(n + 5) is a multiple of
MCQ 131 Mark
By principle of mathematical induction, $2^{4 n-1}$ is divisible by which of the following?
Answer
- 8
Solution:
$P(n)=2^{4 n-1} P(1)=2^3=8$
Let us assume $P(k)$ is divisible by 8 and can be written as 8 c , 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, $\mathrm{P}(\mathrm{k}+1)$ 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:
MCQ 151 Mark
For each $n \mathrm{~N} \in, 10^{2 \mathrm{n}-1}+1$ is divisible by
View full question & answer→MCQ 161 Mark
If Sn is divisible for every n, then Sn is:
MCQ 171 Mark
Let $S(k)=1+3+5+(2 k-1)=3+K^2$. 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
View full question & answer→MCQ 191 Mark
For n ∈ 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
- 3
Solution:
$P(n)=n^3+5 n$
$P(1)=1+5$
$P(1)=6$
We assume the $P(k)$ is true and divisible by 6 .
$P(k)=k^3+5 k$ is divisible by 6 and can be written as 6 c or $3 \times 2 c$
We need to prove that $P(k+1)$ is divisible by 6
$P(k+1)=(k+1)^3+5(k+1)$
$P(k+1)=k^3+1+3 k^2+3 k+5 k+5$
$P(k+1)=\left(k^3+5 k\right)+3 k^2+3 k+6$
$P(k+1)=6 c+3\left(k^2+k+2\right)$
$P(k+1)=(3 \times 2 c)+3\left(k^2+k+2\right)$
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 ∈ N
AnswerLet P(n): n(n + 1)(n + 5) is a multiple of 3.
For n = 1, the given expression becomes (1 × 2 × 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
⇒ 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
⇒ P(k + 1) (k + 1 ) (k + 2) (k + 6) is a multiple of 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 ∈ N
View full question & answer→MCQ 221 Mark
If $\forall \mathbf{m} \in \mathbf{N}$, then $11^{\mathrm{m}+2}+12^{2 \mathrm{~m}-1}$ is divisible by:
Answer
- None of these
Solution:
To find the divisor of $11^{\mathrm{m}+2}+12^{2 \mathrm{~m}-1}$ 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 $11^{m+2}+12^{2 m-1}$ 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
Answer
- True
Solution:
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, $\mathrm{P}(\mathrm{n})$ 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 $\mathrm{n}=1$
It is given that $P(k)$ is true, so $P(k)$ is true for $n=k$
$P(K) \Rightarrow 3^k>k^k$
Now,
$P(k+1)=3^{k+1}>(k+1)^{k+1}$
$3^k \cdot 3^1>(k+1)^k(k+1)$
By the principle for mathematics production,
$\mathrm{P}(\mathrm{n}+1)$ is true, when $\mathrm{P}(\mathrm{n})$ is true
$\therefore \mathrm{P}(\mathrm{n}+1)$ is true. View full question & answer→MCQ 241 Mark
For any natural number n, $7^n-2^n$ is divisible by
Answer
- 5
Solution:
Given, $7^n-2^n$
Let $\mathrm{n}=1$
$7^n-2^n=7^1-2^1=7-2=5$
which is divisible by 5
Let $\mathrm{n}=2$
$7^n-2^n=72-22=49-4=45$
which is divisible by 5
Let $\mathrm{n}=3$
$7^n-2^n=7^3-2^3=343-8=335$
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?
Answer
- (mn)k = mknk
Solution:
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, $\mathrm{P}(\mathrm{n})$ is true for all natural numbers n
Calculation:
Given:
For, $k=1$
Option 1,
$(m n)^1=m^1 n^1, m n=m n($ LHS $=$ RHS) hence it is true
Let us assume that the statement is true for, $k=p$
$(m n)^p=m^p n^p$
Multiplying the above equation with mn , we get mn we get,
$(m n) p+1=m p+1 n p+1$
$m p+1 n p+1=m p+1 n p+1$
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 $n \in N, 2.4^{2 n+1}+3^{3 n+1}$ is divisible by
View full question & answer→MCQ 271 Mark
For each $n \in N, 3^{2 n}-1$ is divisible by:
View full question & answer→MCQ 281 Mark
$10 n+3\left(4^{n+2}\right)+5$ 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. $\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{2}$
- $\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{2}$
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 ϵ 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:
View full question & answer→MCQ 331 Mark
For each n ∈N, the correct statement is
MCQ 341 Mark
Let $P(n)=5^n-2^n \cdot P(n)$ 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
- 1
Solution:
$5^n-2^n$ 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
$\left(2 \cdot 7^{\mathrm{N}}+3 \cdot 5^{\mathrm{N}}-5\right)$ is divisible by ……….. for all $\mathrm{N} \in \mathrm{N}$ :
Answer
- 24
Solution:
Let $P(n):\left(2 \cdot 7^n+3 \cdot 5^n-5\right)$ is divisible by 24 .
For $\mathrm{n}=1$, the given expression becomes $\left(2 \cdot 7^1+3 \cdot 5^1-5\right)=24$, which is clearly divisible by 24 .
So, the given statement is true for $\mathrm{n}=1$, i.e., $\mathrm{P}(1)$ is true.
Let $P(k)$ be true. Then,
$\mathrm{P}(\mathrm{k}):\left(2 \cdot 7^{\mathrm{n}}+3 \cdot 5^{\mathrm{n}}-5\right)$ is divisible by 24 .
$\Rightarrow\left(2 \cdot 7^n+3 \cdot 5^n-5\right)=24 m$, for $m=N$
Now, $\left(2 \cdot 7^n+3 \cdot 5^n-5\right)$
$=\left(2 \cdot 7^{\mathrm{k}} \cdot 7+3 \cdot 5^{\mathrm{k}} \cdot 5-5\right)$
$=7\left(2 \cdot 7^k+3 \cdot 5^k-5\right)=6 \cdot 5^k+30$
$=(7 \times 24 \mathrm{~m})-6\left(5^{\mathrm{k}}-5\right)$
$=(24 \times 7 \mathrm{~m})-6 \times 4 \mathrm{p}$, where $\left(5^{\mathrm{k}}-5\right)=5\left(5^{\mathrm{k}-1}-1\right)=4 \mathrm{p}$
[Since $\left(5^{\mathrm{k}-1}-1\right)$ is divisible by $(5-1)$ ]
$=24 \times(7 \mathrm{~m}-\mathrm{p})$
$=24 \mathrm{r}$, where $\mathrm{r}=(7 \mathrm{~m}-\mathrm{p}) \in \mathrm{N}$
$\Rightarrow P(k+1):\left(2 \cdot 7^k+13 \cdot 5^k+1-5\right)$ is divisible by 24 .
$\Rightarrow P(k+1)$ is true, whenever $P(k)$ is true.
Thus, $\mathrm{P}(1)$ is true and $\mathrm{P}(\mathrm{k}+1)$ is true, whenever $\mathrm{P}(\mathrm{k})$ 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 $\mathrm{P}(\mathrm{n}):$ " $2^{\mathrm{n}}<(1 \times 2 \times 3 \times \ldots \times n)$". Then the smallest positive integer for which $\mathrm{P}(\mathrm{n})$ is true is:
Answer
- 4
Solution:
$\mathrm{P}(1)$ : $2^1<1$
$2<1$ is false
$\mathrm{P}(2)$ : $2^2<1 \times 2$
$4<2$ is false
$\mathrm{P}(3): 2^3<1 \times 2 \times 3$
$8<6$ is false
$\mathrm{P}(4): 2^4<1 \times 2 \times 3 \times 4$
$16<24$ 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
View full question & answer→MCQ 381 Mark
Consider the statement: “P(n) : n2 – 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.
- 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 n Î N, then the least positive integral value of k is:
MCQ 431 Mark
Let P(n) = n(n + 1) is an even number, then which of the following satisfy P(n):
AnswerGiven, P(n) = n(n + 1)
$\therefore$ P(3) = 3.4 = 12(even)
P(100) = 100 × 101 = 10100(even)
P(50) = 50 × 51 = 2520(even)
As P(3), P(100), P(50) are even numbers.
Short cut Method: P(n) = n(n + 1)
Product of two consecutive integer (natural numbers) is always even.
View full question & answer→MCQ 441 Mark
For all $\mathrm{n} \in \mathrm{N}, 41^{\mathrm{n}}-14^{\mathrm{n}}$ 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 $13+23+33+\ldots \ldots . .+n^3$ ?
- 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. $\Big(\frac{\text{n}(\text{n+1)}}{2}\Big)^2$
- $\Big(\frac{\text{n}(\text{n+1)}}{2}\Big)^2$
View full question & answer→MCQ 471 Mark
If $P(k)=k^2(k+3)\left(k^2-1\right)$ is true, then what is $P(k+1) ?$
AnswerCorrect option: C. $(k+1)^2(k+4) k(k+2)$
- $(k+1)^2(k+4) k(k+2)$
Solution:
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 $k \cdot P(k+1)=(k+1)^2(k+1+3)\left((k+1)^2-1\right)$
$P(k+1)=(k+1)^2(k+4)\left(k^2+1+2 k-1\right)$
$P(k+1)=(k+1)^2(k+4)\left(k^2+2 k\right)$
$P(k+1)=(k+1)^2(k+4) k(k+2)$ View full question & answer→MCQ 481 Mark
For all $n \in N, 3 n^5+5 n^3+7 n$ is divisible by:
Answer
- 15
Solution:
Given number $=3 n^5+5 n^3+7 n$
Let $\mathrm{n}=1,2,3,4$, ........
$3 n^5+5 n^3+7 n=3 \times 1^2+5 \times 1^3+7 \times 1=3+5+7=15$
$3 n^5+5 n^3+7 n=3 \times 2^5+5 \times 2^3+7 \times 2=3 \times 32+5 \times 8+7 \times 2=96+40+14=150=15 \times 10$
$3 n^5+5 n^3+7 n=3 \times 3^5+5 \times 3^3+7 \times 3=3 \times 243+5 \times 27+7 \times 3=729+135+21=885=15 \times 59$
Since, all these numbers are divisible by 15 for $\mathrm{n}=1,2,3, \ldots .$.
So, the given number is divisible by 1 View full question & answer→MCQ 491 Mark
If $n \in N$, then $n\left(n^2-1\right)$ is divisible by:
Answer
- 6
Solution:
$n\left(n^2-1\right)=n(n-1)(n+1)$
One of the $n, n+1$ and $n-1$ will be a multiple of 3 .
Since $n-1, n$ and $n+1$ are three consecutive integers, therefore at least one of them will be divisible by 2 .
Therefore $n\left(n^2-1\right)$ is divisible by 6 . View full question & answer→MCQ 501 Mark
For every positive integer $n, n 7 / 7+n 5 / 5+2 n^3 / 3-n / 105$ is:
View full question & answer→MCQ 511 Mark
$x\left(x^{n-1}-n \alpha^{n-1}\right)+\alpha^n(n-1)$ is divisible by $(x-\alpha)^2$ for:
View full question & answer→MCQ 521 Mark
Let T(k) be the statement 1 + 3 + 5 + .... + (2k – 1)= k +10 Which of the following is correct:
- A
- ✓
T(k) is true ⇒ T(k + 1) is true
- C
- D
AnswerCorrect option: B. T(k) is true ⇒ T(k + 1) is true
- T(k) is true ⇒ T(k + 1) is true
View full question & answer→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:
AnswerFor 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.
View full question & answer→MCQ 541 Mark
If $n \in N$, then $11^{n+2}+12^{2 n+1}$ is divisible by:
View full question & answer→MCQ 551 Mark
1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + n(n + 1):
AnswerLet 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)}\
⇒ P(k + 1): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +……+ (k + 1) (k + 2)
= (1/3){k + 1 )(k + 2) (k +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 values of ∈ N
View full question & answer→MCQ 561 Mark
$\forall n \in N ; x^{2 n-1}+y^{2 n-1}$ is divisible by?
- A
$x-y$
- ✓
$x+y$
- C
$x y$
- D
$x^2+y^2$
View full question & answer→MCQ 571 Mark
If P(n): “46 + 16 + k is divisible by 64 for n Î N” is true, then the least negative integral value of k is:
MCQ 581 Mark
For all $10^n+3\left(4^{n+2}\right)+5$ is divisible by $(n \in N)$ :
View full question & answer→MCQ 591 Mark
If $n \in N, 7^{2 n} 48^{n-1}$ - is divisible by:
View full question & answer→MCQ 601 Mark
Let $x>-1$, then statement $p(n):(1+x)^n>1+n x$, where $n \in N$ is true for
- A
- B
- ✓
For all n > 1, provided x = 0.
- D
AnswerCorrect option: C. For all n > 1, provided x = 0.
- For all n > 1, provided x = 0.
Solutions:
$p(1):(1+x)^1>1+x$ 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 $x=0$ View full question & answer→MCQ 611 Mark
$P(n): 2 \times 7n + 3 \times 5n - 5$ is divisible by:
Answer
- 24, ∀ n ∈ N
Solution:
Concepts:
Suppose there is a given statement $\mathrm{P}(\mathrm{n})$ 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 $\mathrm{n}=\mathrm{k}$ (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, $\mathrm{P}(\mathrm{n})$ is true for all natural numbers n
Calculation:
Given:
$P(n)=2 \times 7 n+3 \times 5 n-5$
Put $\mathrm{n}=1$
$P(1)=2 \times 7^1+3 \times 5^1-5=24$, Which is divisible by 24
Assume $P(k)$ is true
$P(k)=2 \times 7^k+3 \times 5^k-5=24 q \text {, where } q \in N$
Now,
$\mathrm{T}(\mathrm{k}+1)=2 \times 7^{\mathrm{k}+1}+3 \times 5^{\mathrm{k}+1}-5=2 \times 7^{\mathrm{k}} \times 7+3 \times 5^k \times 5-5$
$\Rightarrow 7\left\{2 \times 7^{\mathrm{k}}+3 \times 5^k-5-3 \times 5^k+5\right\}+3 \times 5^k \times 5-5$
$\Rightarrow 7\left\{24 \mathrm{q}-3 \times 5^k+5\right\}+15 \times 5^k-5$
$\Rightarrow(7 \times 24 q)-21 \times 5^k+35+15 \times 5^k-5$
$\Rightarrow(7 \times 24 q)-6 \times 5^k+30=(7 \times 24 q)-6\left(5^k-5\right)$
$\Rightarrow(7 \times 24 q)-6 \times(4 p)\left\{\text { As }\left(5^k-5\right) \text { is a multiple of } 4\right\}$
$\Rightarrow(7 \times 24 q)-24 p=24(7 q-p)$
$\Rightarrow 24 \times r, r=7 q-p, \text { is some natural number ---(2) }$
Thus, $P(k+1)$ is true whenever $P(k)$ is true View full question & answer→MCQ 621 Mark
For each natural number, the statement $P(n)=2^{3 n}-1$ is divisible by:
Answer
- 7
Solution:
$P(n)=2^{3 n}-1=8^n-1$
$P(n)=(1+7)^n-1$
$\Rightarrow P(n)=1+{ }^n C_1 \times 7+{ }^n C_2 \times 7^2+\ldots+{ }^n C_n \times 7^n-1$
$\Rightarrow P(n)=7\left({ }^n C_1+{ }^n C_2 7+\ldots+{ }^n C_n 7^{n-1}\right)$
Therefore, $\mathrm{P}(\mathrm{n})$ is divisible by 7 . View full question & answer→MCQ 631 Mark
Consider the following statements:1) ~ (p ∧ q) = ~ p ∨ ~ q 2) ~ (p ∨ q) = ~ p ∧ ∼ q 3) ~ (~ p) = p Which of the above statements is/are correct?
AnswerConcept:
- The negation of a conjunction p ∧ q is the disjunction of the negation of p and the negation of q. Equivalently, we write ~ (p ∧ q) = ~ p ∨ ~ q
- The negation of a disjunction p ∨ q is the conjunction of the negation of p and the negation of q. Equivalently, we write ~ (p ∨ q) = ~ p ∧ ∼ q
- Negation of negation of a statement is the statement itself. Equivalently, we write ~ (~ p) = p
View full question & answer→MCQ 641 Mark
For each $n \in N$, then $3^{2 n+1}+1$ is divisible by:
View full question & answer→MCQ 651 Mark
If n ∈ N $\Big(\frac{\text{n}+1}{2}\Big)^\text{n}\geq\text{n!}$ is true when:
AnswerConcept:
$\text{n!}=\text{n}\times(\text{n}-1)\times\text{n}-2.....\times3\times2\times1$
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
View full question & answer→MCQ 661 Mark
For all positive integral values ofn, $3^2 n-2 n+1$ is divisible by:
View full question & answer→MCQ 671 Mark
If P(n) be the statement n(n + 1) + 1 is odd, then which of the following is false?
AnswerP(n) = n(n + 1) + 1
P(2) = 6 + 1 = 7
P(3) = 3 × 4 + 1 = 13
P(4) = 4 × 5 + 1 = 21
None of the above is even.
View full question & answer→MCQ 681 Mark
n(n + 1)(n + 5) is a multiple of 3 is true for:
- A
- B
Only natural number 3 ≤ n < 15
- ✓
- D
View full question & answer→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:
Answer
- -1
Solution:
Given that $P(n): 49^n+16^n+k$ is divisible by 64 for $n \in N$
For $\mathrm{n}=1$,
$P(1): 49+16+k=65+k$ is divisible by 64 .
Thus $k$, should be -1 since, $65-1=64$ is divisible by 64 . View full question & answer→MCQ 701 Mark
For all $n \in N, 3.5^{2 n+1}+2^{3 n+1}$ is divisible by:
View full question & answer→MCQ 711 Mark
If n is an odd positive integer, then aⁿ + bⁿ is divisible by:
AnswerGiven number = aⁿ + bⁿ
Let n = 1, 3, 5, ……..
aⁿ + bⁿ = a + b
aⁿ + bⁿ = a³ + b³ = (a + b) × (a² + b² + 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)
View full question & answer→MCQ 721 Mark
For all $n \geq 2, n^2\left(n^4-1\right)$ is divisible by:
View full question & answer→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
AnswerGiven,
$10\text{n }+3\times^{\text{n}+2}+\lambda$ 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.
View full question & answer→MCQ 741 Mark
For all n∈N, 72n − 48n−1 is divisible by:
Answer
- 2304
Solution:
Concepts:
Suppose there is a given statement $\mathrm{P}(\mathrm{n})$ 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:
$P(n)=72 n-48 n-1$
Put, $n=1$
$P(1)=72-48 \times 1-1=0$
Check the expression $\mathrm{P}(\mathrm{n})$ for $\mathrm{n}=\mathrm{k}$ (where k is some positive integer) $=2,3,4$.....
$P(2)=7^{2 n}-48 n-1=7^4-48 \times 2-1=2401-96-1=2401-97=2304$
$P(3)=7^{2 n}-48 n-1=7^6-48 \times 3-1=117649-144-1=117649-145=117504=2304 \times 51$
Since, all these numbers are divisible by 2304 for $\mathrm{n}=1$ and $\mathrm{k}=2,3, \ldots \ldots$
So, the given number is divisible by 2304 View full question & answer→MCQ 751 Mark
Let $P(n)$ be the statement that $n^2-n+41$ is prime, then which of the following is not true?
Answer
- P(41)
Solution:
given that $P(n)=n^2-n+41$
$\mathrm{P}(2)=2^2-2+41=43$ (prime true)
$P(3)=3^2-3+41=47$ (prime true)
$P(41)=41^2-41+41=(41)^2$
$\therefore \mathrm{P}(41)$ is not true. View full question & answer→MCQ 761 Mark
What is the sum of 2 + 4 + 6 + 8 +....+ 2n ? A:
MCQ 771 Mark
If $x^n-1$ is divisible by $x-k$, then the least positive integral value of $k$ is:
Answer
- 1
Solution:
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 View full question & answer→MCQ 781 Mark
The smallest positive integer n for which $ \text{n}!<\Big(\frac{\text{n}+1}{2}\Big)\text{n}$ holds, is:
View full question & answer→MCQ 791 Mark
72n + 16n - 1 is divisible by (n ∈ N):
AnswerCalculation:
S = 72n + 16n - 1
For n = 1
S = 49 + 16 - 1 = 64
For n = 2
S = 2401 + 32 - 1 = 2432 = 64 × 32, which is divisible by 64
View full question & answer→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!
AnswerCorrect 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!
View full question & answer→MCQ 811 Mark
If P(n): 3n < n!, n ϵ N, Then P(n) is true for:
Answer
- n ≥ 7
Solution:
Concept:
$\text{n}!=\text{n}\times(\text{n}-1)\times(\text{n}-2).....\times3\times2\times1$
Calculation:
Given:
$P(n): 3 n<n!$
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 $n=3$
$\mathrm{P}(\mathrm{n})=3 \mathrm{n}<\mathrm{n!}=3^3<3!$
$\Rightarrow 27</ 3 \times 2 \times 1,27<6 \text {, Hence Option } 2 \text { is wrong }$
Putting $n=6$
$P(n)=3 n<n!=63<6!$
$P(n)=3 n<n!=3^6<6!, 1029<6 \times 5 \times 4 \times 3 \times 2 \times 1$
$1029<720$ Hence Option 3 is wrong
Put $\mathrm{n}=7$
$P(n)=3 n<n!=3^7<7!, 2187<7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
$2187<5040$, Option 1 satisfies the given expression, Hence the correct answer is option 1. View full question & answer→MCQ 821 Mark
For all $n \in N, 5^{2 n}-1$ is divisible by:
Answer
- 24
Solution:
Given number $=5^{2 n}-1$
Let $\mathrm{n}=1,2,3,4, \ldots \ldots \ldots$.
$5^{2 n}-1=5^2-1=25-1=24$
$5^{2 n}-1=5^4-1=625-1=624=24 \times 26$
$5^{2 n}-1=5^6-1=15625-1=15624=651 \times 24$
Since, all these numbers are divisible by 24 for $n=1,2,3, \ldots$.
So, the given number is divisible by 2 View full question & answer→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,
View full question & answer→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 ∈ N and also that P(5) is true. Based on this, he could conclude that P(n) is true:
AnswerThe student could be able to conclude that P(n) is true for all n ≥ 5 since P(5) is true for all k > 5 ∈ N as well as true for P(5) and P(k + 1) is true, whenever P(k) is true.
View full question & answer→MCQ 851 Mark
For positive integer $n, 10^{n-2}>81 n$, if:
View full question & answer→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:
Answer
- 1
Solution:
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 . View full question & answer→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
- ✓
- D
Answer
- All natural numbers n
Solutions:
Let the statement be denoted by $p(n)$ i.e.,
$P(n): n(n+1)(n+5)$ is a multiple of 3
For $n=1, n(n+1)(n+5)=1.2 .6=12=3.4$
$\mathrm{P}(\mathrm{n})$ is true for $\mathrm{n}=1$
Suppose $p(k)$ is true for $\mathrm{n}=\mathrm{k}$ i.e.
$k(k+1)(k+5)=3 m \text { (let) or } k 3+6 k 2+5 k=3 m \ldots \text {... (i) }$
Replacing $k$ by $k+1$, we get
$(k+1)(k+2)(k+6)=k\left(k^2+8 k+12\right)+\left(k^2+8 k+12\right)$
$k^3+9 k^2+20 k+12=\left(k^3+6 k^2+5 k\right)+\left(3 k^2+15 k+12\right)$
$=3 m+3 k^2+15 k+12[\text { from }(i)]$
$=3\left(m+k^2+5 k+4\right)$
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 $n \in N$
View full question & answer→MCQ 881 Mark
The greatest positive integer, which divides n(n +1)(n + 2)(n + 3) for all n Î N, is:
MCQ 891 Mark
If $n \in N, 7^{2 n}-48 n-1$ is divisible by:
View full question & answer→MCQ 901 Mark
If an = √7 + √7 + √7 +... ... 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
$\text{an}<3"\text{n}\geq1$
AnswerCorrect option: B. $\text{an}<7"\text{n}\geq1$
- $\text{an}<7"\text{n}\geq1$
View full question & answer→MCQ 911 Mark
The sum of the series 1³ + 2³ + 3³ + ………..n³ is:
- A
$\Big(\frac{\text{(n}+1)}{2}\Big)^2$
- B
$\Big(\frac{\text{n}}{2}\Big)^2$
- C
$\Big(\frac{\text{n}\text{(n}+1)}{2}\Big)$
- ✓
$\Big(\frac{\text{n}\text{(n}+1)}{2}\Big)^2$
AnswerCorrect option: D. $\Big(\frac{\text{n}\text{(n}+1)}{2}\Big)^2$
Given, series is 1³ + 2³ + 3³ + ……….. n³
$\Big(\frac{\text{n}\text{(n}+1)}{2}\Big)^2$
View full question & answer→MCQ 921 Mark
Choose the correct answer. For all $n \in N, 3 \times 5^{2 n+1}+2^{3 n+1}$ is divisible by:
Answer
- 17
Solution:
Let $\text{P(n): }3\times5^{2\text{n}+1}+2^{3\text{n}+1}$
For $\text{P}(1):\ 3\times5^{2\times1+1}+2^{3\times1+1}=3.5^3+2^4$
$=3\times(125)+16=375+16$
$=391=23\times17$
So, P(1) is divisible by 17.
Assume P(k) is divisible by $17\Rightarrow\text{P(k): }3\times5^{2\text{k}+1}+2^{3\text{k}+1}=17\lambda_1,\lambda_1\in\text{N}$
So, $3\times5^{2\text{k}+1}=17\lambda_1-2^{3\text{k}+1}\ ....(\text{i})$
Now we have to prove P(k + 1) is divisible by $17\Rightarrow\text{P(k+1): }3\times5^{2\text{k}+3}+2^{3\text{k}+4}=17\lambda_2,\lambda_2\in\text{N}$
$=\big[17\lambda_1-2^{3\text{k}+1}\big]\times5^2+2^{3\text{k}+1}\cdot2^{3}$
$=17\lambda_1\cdot25-25\cdot2^{3\text{k}+1}+8\cdot2^{3\text{k}+1}$
$=17\lambda_1\cdot25-17\cdot2^{3\text{k}+1}$
$=17\big[25\lambda_1-2^{3\text{k}+1}\big]$
$=17\lambda_2,\lambda_2=25\lambda_1-2^{3\text{k}+1}$
So, P(k + 1) is divisible by 17 whenever P(k) is divisible by 17.
Hence by the principle of mathematical induction we have $P(n): 3 \times 5^{2n+1} + 2^{3n+1}$ is divisible by $17,\forall\text{n}\in\text{N}$ View full question & answer→MCQ 931 Mark
For n∈ N,$\big(\frac{1}{5}\big)\text{n}^5+\big(\frac{1}{3}\big)\text{n}^3+\big(\frac{1}{15}\big)$ is:
View full question & answer→MCQ 941 Mark
$P(n)=n\left(n^2-1\right)$. Which of the following does not divide $P(k+1)$ ?
Answer
- k + 3
Solution:
$P(n)=n\left(n^2-1\right)$
$P(k+1)=(k+1)\left((k+1)^2-1\right)$
$P(k+1)=(k+1)\left(k^2+1+2 k-1\right)$
$P(k+1)=(k+1)\left(k^2+2 k\right)$
$P(k+1)=(k+1) k(k+2)$
Therefore, $k,(k+1),(k-1)$ divide $P(k+1)$. View full question & answer→MCQ 951 Mark
The greatest positive integer, which divides (n + 2) (n + 3) (n + 4) (n + 5) (n + 6) for all n ∈ N, is:
MCQ 961 Mark
$\mathrm{n}^2<2^{\mathrm{n}}$ for all natural numbers:
Answer
- n ≥ 5
Solution:
Consider, $\mathrm{P}(\mathrm{n}): \mathrm{n}^2<2^{\mathrm{n}}$
Substituting $\mathrm{n}=1,2,3, \ldots$
$P(1): 1^2<2^1$
$1<2 \text { (not true) }$
$P(2): 2^2<2^2$
$4<4 \text { (not true) }$
$P(3): 3^2<2^3$
$9<8 \text { (not true) }$
$P(4): 4^2<2^4$
$16<16 \text { (not true) }$
$P(5): 5^2<2^5$
$25<32 \text { (true) }$
$P(6): 6^2<2^6$
$26<64 \text { (true) }$
Thus, $\mathrm{n}^2<2^n$ for all natural numbers $\mathrm{n} \geq 5$. View full question & answer→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:
View full question & answer→MCQ 981 Mark
Let P(n) be a statement and P(n) = P(n + 1)∀n ∈ N, then P(n) is true for what values of n?
- ✓
- B
- C
For all n > m , m being a fixed positive integer
- D
AnswerGiven, P(n) = P(n+1)∀n ∈ N
Substituting n - 1 in place of n,
P(n - 1) = P(n)
Thus if P(k) is true for some k ∈ N, then it is true for k - 1 and k + 1.
Thus, it is true ∀k ∈ N
View full question & answer→MCQ 991 Mark
$x\left(x^{n-1}-n \alpha^{n-1}\right)+\alpha^n(n-1)$ is divisible by $(x-\alpha)^2$ for:
View full question & answer→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.
AnswerCorrect 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.
View full question & answer→MCQ 1011 Mark
For all n Î N, 3.52n + 1 + 23n + 1 is divisible by:
MCQ 1021 Mark
$2.4^{2\text{n+1}}+3^{3\text{n+1}}$ is divisible by: (for all n ∈ N)
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., 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. View full question & answer→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:
MCQ 1041 Mark
If $x^n-1$ is divisible by $x-\lambda$, then the least positive integral value of $\lambda$ is:
Answer
- 1
Solution:
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 $\boldsymbol{\lambda}$ is 1. View full question & answer→MCQ 1051 Mark
By mathamatical induction $n\left(n^2-1\right)$ is divisible by:
View full question & answer→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:
View full question & answer→MCQ 1071 Mark
n(n + 1) (n + 5) is a multiple of:
AnswerLet 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
View full question & answer→MCQ 1081 Mark
If n ∈ N, then 121n - 25n + 1900n - (-4) n is divisible by which one of the following?
Answer
- 2000
Solution:
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. View full question & answer→MCQ 1091 Mark
If $\mathrm{n} \in \mathrm{N}$, then $11^{\mathrm{n}+2}+12^{2 \mathrm{n}+1}$ is divisible by:
View full question & answer→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:
View full question & answer→MCQ 1111 Mark
Mathematical Induction is the principle containing the set.
AnswerMathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers.
View full question & answer→MCQ 1121 Mark
For all $\mathrm{n} \in \mathrm{N}, 3^{2 \mathrm{n}}+7$ is divisible by:
Answer
- 8
Solution:
Given number $=32 \mathrm{n}+7$
Let $\mathrm{n}=1,2,3,4, \ldots \ldots .$.
$3^{2 n}+7=3^2+7=9+7=16$
$3^{2 n}+7=3^4+7=81+7=88$
$3^{2 n}+7=3^6+7=729+7=736$
Since, all these numbers are divisible by 8 for $\mathrm{n}=1,2,3, \ldots .$.
So, the given number is divisible by 8 View full question & answer→MCQ 1131 Mark
For every integer $\mathbf{n} \geq 1,\left(3^{2 n}-1\right)$ is always divisible by:
- A
$2^{n 2}$
- B
$2^{n+4}$
- ✓
$2^{n+2}$
- D
$2^{n+3}$
AnswerCorrect option: C. $2^{n+2}$
View full question & answer→MCQ 1141 Mark
The sum of n terms of the series 1² + 3² + 5² +……… is:
AnswerLet S = 1² + 3² + 5² +………(2n – 1)²
⇒ S = {1² + 2² + 3² + 4² ………(2n – 1)² + (2n)²} – {2² + 4² + 6² +………+ (2n)²}
⇒ S = {2n × (2n + 1) × (4n + 1)}/6 – {4n × (n + 1) × (2n + 1)}/6
⇒ S = n(4n² – 1)/3
View full question & answer→MCQ 1151 Mark
For each $n \mathrm{~N} \in, 3^{2 \mathrm{n}-1}$ is divisible by:
View full question & answer→MCQ 1161 Mark
For all n ∈ N ,n (n+1 )(n+5 ) is a multiple of:
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
$2^{\text{n + 3}}$
AnswerCorrect 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.
View full question & answer→MCQ 1181 Mark
For every positive integer $n, 7n – 3n$ is divisible by
Answer
- 4
Solution:
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, $\mathrm{P}(\mathrm{n})$ is true for all natural numbers n
Calculation:
We have to find $7^n-3^n$ is divisible by which number
Consider $\mathrm{P}(\mathrm{n})$ : $7 \mathrm{n}-3 \mathrm{n}$
$P(1): 7^1-3^1=4$
Thus, $7 n-3 n$ is divisible by 4
Let $P(k)$ is true for $n=K$
$\Rightarrow 7^{\mathrm{k}}-3^{\mathrm{k}}$ is divisible by 4
So, $7 n-3 n=4 d$
Now, prove that $\mathrm{P}(\mathrm{k}+1)$ is true.
$\Rightarrow 7^{(k+1)}-3^{(k+1)}=7^{(k+1)}-7.3^k+7.3^k-3^{(k+1)}$
$=7\left(7^k-3^k\right)+(7-3) 3^k$
$=7(4 d)+(7-3) 3^k$
$=7(4 d)+4.3^k$
$=4\left(7 \mathrm{~d}+3^k\right)$
Hence, $P(n): 7^n-3^n$ is divisible by 4 is true. View full question & answer→MCQ 1191 Mark
If $\forall \mathrm{m} \in \mathrm{N}$, then $11^{\mathrm{m}+2}+12^{2 \mathrm{~m}-1}$ is divisible by:
View full question & answer→MCQ 1201 Mark
$3+13+29+51+79+\ldots$ to n terms $=:$
- A
$2 n^2+7 n^3$
- B
$n^2+5 n^3$
- ✓
$n^3+2 n^2$
- D
AnswerCorrect option: C. $n^3+2 n^2$
View full question & answer→MCQ 1211 Mark
For all positive integral values of $n, 3^{2 n}-2 n+1$ is divisible by:
View full question & answer→MCQ 1221 Mark
$\mathrm{n}^2+3 \mathrm{n}$ is always divisible by which number, provided n is an integer?
Answer
- 2
Solution:
$P(n)=n^2+3 n$
$P(1)=1+3$
$P(1)=4$
Let's assume that $P(k)$ is true and divisible by 4 . Therefore, $P(k)=k^2+3 k$ can be written as $4 c$.
We need to check if $P(k+1)$ is divisible by 4
$\mathrm{P}(\mathrm{k}+1)=(\mathrm{k}+1)^2+3(\mathrm{k}+1)$
$\mathrm{P}(\mathrm{k}+1)=\mathrm{k}^2+1+2 \mathrm{k}+3 \mathrm{k}+3$
$\mathrm{P}(\mathrm{k}+1)=\mathrm{k}^2+5 \mathrm{k}+4$
$\mathrm{P}(\mathrm{k}+1)=\left(\mathrm{k}^2+3 \mathrm{k}\right)+2 \mathrm{k}+4$
$\mathrm{P}(\mathrm{k}+1)=4 \mathrm{c}+2 \mathrm{k}+4$
$\mathrm{P}(\mathrm{k}+1)=4 \mathrm{c}+2(\mathrm{k}+2)$
Clearly the second part of the equation is not divisible by 4 . However $P(k)=4 c$ is divisible by 2 and $P(k+1)$ is also divisible by 2 . Therefore, 2 divides $P(n)$ View full question & answer→MCQ 1231 Mark
If $x^{2 n-1}+y^{2 n-1}$ is divisible by $x+y$, if $n$ is:
View full question & answer→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
$k^4+3 k^3+9 k^2+6 k+2$
AnswerCorrect option: B. $k^4+5 k^3+9 k^2+7 k+2$
- $k^4+5 k^3+9 k^2+7 k+2$
Solution:
$P(n)=n^3(n+1)$
$P(k+1)=(k+1)^3(k+1+1)$
$P(k+1)=\left(k^3+3 k^2+3 k+1\right)(k+2)$
$P(k+1)=k^4+3 k^3+3 k^2+k+2 k^3+6 k^2+6 k+2$
$P(k+1)=k^4+5 k^3+9 k^2+7 k+2$ View full question & answer→MCQ 1251 Mark
For all $n ∈ N, 3n^5 + 5n³ + 7n$ is divisible by:
Answer
- 15
Solution:
Given number $=3 n^5+5 n^2+7 n$
Let $n=1,2,3,4$,........
$3 n^5+5 n^3+7 n=3 \times 1^2+5 \times 1^3+7 \times 1=3+5+7=15$
$3 n^5+5 n^3+7 n=3 \times 2^5+5 \times 2^3+7 \times 2=3 \times 32+5 \times 8+7 \times 2=96+40+14=150=15 \times 10$
$3 n^5+5 n^3+7 n=3 \times 3^5+5 \times 3^3+7 \times 3=3 \times 243+5 \times 27+7 \times 3=729+135+21=885=15 \times 59$
Since, all these numbers are divisible by 15 for $n=1,2,3, \ldots .$.
So, the given number is divisible by 15 View full question & answer→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:
AnswerAs 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.
View full question & answer→MCQ 1271 Mark
For all $\text{n}\in\text{N}, 3 \times 5^{2n+1} + 2^{3n+1}$ is divisible by:
Answer
- 17
Solution:
$3.5^{2n+ 1} + 2^{3n+1}$ 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. View full question & answer→MCQ 1281 Mark
The product of three consecutive natural numbers is divisible by:
- $3$
- $8$
- $6$
- $11$
- ✓
$a$ and $c$
- B
$b$ and $c$
- C
$c$ and $d$
- D
$a$ and $d$
AnswerCorrect option: A. $a$ and $c$
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,
$P(1) = 1 \times 2 \times 3 = 6,$ which is divisible by $3$ and $6.$
$P(2) = 2 \times 3 \times 4 = 24,$ which is divisible by $3, 8$ and $6.$
$P(3) = 3 \times 4 \times 5 = 60,$ which is divisible by $3$ and $6.$
$P(4)= 4 \times 5 \times 6 = 120,$ which is divisible by $3, 8$ and $6.$
Hence, $P(n)$ is divisible by $3$ and $6$ for all $n \in N.$
View full question & answer→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:
Answer
- 9
Solution:
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 $n \in N$. View full question & answer→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:
View full question & answer→MCQ 1311 Mark
Ifn ∈ N, then the highest positive integerwhich dividesn(n – 1)(n – 2) is:
MCQ 1321 Mark
Let P(n): “2n < (1 × 2 × 3 × ... × n)”. Then the smallest positive integer for which P(n) is true is:
MCQ 1331 Mark
$(n^2 + n)$ is ________ for all $n ∈ N$.
Answer
- Even
Solution:
Concept:
Suppose there is a given statement $\mathrm{P}(\mathrm{n})$ 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, $\mathrm{P}(\mathrm{n})$ is true for all natural numbers n
Calculation:
Given:
$P(n)=n^2+n$
Put, $\mathrm{n}=1$
$P(1)=12+1=2 \text { (Even) }$
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), $P(k+1)=2 m+2(k+1)=2[m+(k+1)]$, 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 $p(n)=\left(n^2+n\right)$ is even View full question & answer→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:
Answer
- 5
Solution:
Given that $10^n+3.4^{n+2}+k$ is exactly divisible by 9 .
Consider: $\mathrm{P}(\mathrm{n})=10^{\mathrm{n}}+3.4^{\mathrm{n}+2}+k$
Substituting $\mathrm{n}=1$,
$P(1)=10^1+3.4^{1+2}+k$
$=10+3(64)+k$
$=10+192+k$
$=202+k$ is exactly divisible by 9 , the value of $k$ will be 5 . View full question & answer→MCQ 1351 Mark
If n is an even number, then the digit in the units place of $2^{2 n}+1$ will be:
Answer
- 7
Solution:
Since $2^{2 n}$ is even therefore $2^{2 n}+1$ is odd, therefore digit at unit place should be odd, rejecting option 3 . Put $\mathrm{n}=2$, we get $2^{2 \mathrm{n}}+1=17$,
Hence digit should be 7 View full question & answer→MCQ 1361 Mark
Let T(k) be the statement 1 + 3 + 5 +...+ (2k – 1) = k + 10 Which of the following is correct?
- A
- ✓
T(k) is true ⇒ T(k + 1) is true
- C
T(n) is true for all n ÎN
- D
AnswerCorrect option: B. T(k) is true ⇒ T(k + 1) is true
- T(k) is true ⇒ T(k + 1) is true
View full question & answer→MCQ 1371 Mark
Let $P(n)=5^n-2^n \cdot P(n)$ 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
- 1
Solution:
$5^n-2^n$ is divisible by $5-2=3$ always... Putting $\mathrm{n}=\lambda=1$ which is the least odd positive integer, this works to be true.
Hence Option C View full question & answer→MCQ 1381 Mark
The sum of the series 1 + 2 + 3 + 4 + 5 +………..n is:
AnswerCorrect option: D. $\frac{\text{n}(\text{n} + 1)}{2}$
Given, series is series 1 + 2 + 3 + 4 + 5 + ………..n
$\text{sum}=\frac{\text{n}(\text{n} + 1)}{2}$
View full question & answer→MCQ 1391 Mark
If n is a positive integer, then 2. 42n + 1 + 33n + 1 is divisible by:
MCQ 1401 Mark
If P(n) is a statement such that P(3) is true. Assuming P(k) is true ⇒ p(k + 1) is true for all $\text{k} \geq 3$, then P(n) is true:
View full question & answer→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:
View full question & answer→MCQ 1421 Mark
The inequality $n!>2^{n-1}$ is true for:
View full question & answer→MCQ 1431 Mark
For all $n \in N, 3.5^{2 n+1}+2^{3 n+1}$ is divisible by:
Answer
- 17
Solution:
Let $P(n)$ be the statement that $3.5^{2 n+1}+2^{3 n+1}$ is divisible by 17
If $\mathrm{n}=1$, then given expression $=3 \times 5^3+2^4+375+16=391=17 \times 23$, divisible by 17 .
$\mathrm{P}(1)$ is true
Assume that $P(k)$ is true.
$3.5^{2 \mathrm{k}+1}+2^{3 \mathrm{k}+1}$ is divisible by 17 .
$3 \cdot 5^{2 \mathrm{k}}=1+2^{3 \mathrm{k}+1}=17 \mathrm{~m}$ where $\mathrm{m} \in \mathrm{N}$
$3.5^{2(k+1)+1}+23^{(k+1)+1}$
$=3.5^{2 \mathrm{k}+1} \times 5^2+2^{3 \mathrm{k}+1} \times 2^3$
$=25^{(17 \mathrm{~m}-23 \mathrm{k}+1)}+8.2^{3 \mathrm{k}+1}$
$=425 \mathrm{~m}-25.2^{3 \mathrm{k}+1}+8.2^{3 \mathrm{k}+1}$
$=425 \mathrm{~m}-17.2^{3 \mathrm{k}+1}$
$=17\left(25 m-2^{3 k+1}\right)$, divisible by 17
$\mathrm{P}(\mathrm{k}+1)$ is true by Principle of Mathematical Induction
$P(n)$ is true for all $n \in N .3 .5^{2 n+1}+2^{3 n+1}$ is divisible by 17 for all $n \in N$ View full question & answer→MCQ 1441 Mark
Let $S(K)=1+3+5 \ldots+(2 K-1)=3+K^2$. Then which of the following is true:
- A
Principle of mathematical induction can be used to prove the formula
- ✓
- C
- D
View full question & answer→MCQ 1451 Mark
If p is a prime number, then np−n is divisible by p for all n, where:
- ✓
- B
- C
N is even natural number.
- D
N is not a composite number.
View full question & answer→MCQ 1461 Mark
If $1+5+12+22+35+...\text{to}\text{ n }\text{terms}=\frac{\text{n}^2(\text{n+1)}}{2},\text{nth }$teram 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
$\frac{\text{n}(4\text{n}+1)}{2}$
AnswerCorrect option: B. $\frac{\text{n}(3\text{n}-1)}{2}$
- $\frac{\text{n}(3\text{n}-1)}{2}$
View full question & answer→MCQ 1471 Mark
$S_n$ is divisible by the multiple of:
View full question & answer→MCQ 1481 Mark
f P(n) is a statement (n ∈ N) such that, if P(k) is true, P(k + 1) is true for k∈ N, then P(n) is true:
MCQ 1491 Mark
The smallest positive integer n for which $\mathrm{n}!<\left(\frac{\mathrm{n}+1}{2}\right)^{\mathrm{n}}$ holds, is:
View full question & answer→MCQ 1501 Mark
If P(n) = 2 + 4 + 6 + .....+ 2n, n Î N , then P(k) = k(k +1) + 2 ⇒ 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:
AnswerCorrect option: D. $\text{none}\text{ of}\text{ these}$
- $\text{none}\text{ of}\text{ these}$
View full question & answer→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:
View full question & answer→MCQ 1521 Mark
{1/(3 ∙ 5)} + {1/(5 ∙ 7)} + {1/(7 ∙ 9)} + ……. + 1/{(2n + 1) (2n + 3)}:
AnswerLet 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 LHS = 1/(3 ∙ 5) = 1/15 and RHS = 1/{3(2 × 1 + 3)} = 1/15.
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)}
= (2k² + 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}]
⇒ 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 ∈ N
View full question & answer→MCQ 1531 Mark
Choose the correct answer. If $10^n + 3 \times 4^{n+2} + k$ is divisible by 9 for all n ∈ N, then the least positive integral value of k is:
Answer
- 5
Solution:
Let $\mathrm{P}(\mathrm{n})=10^{\mathrm{n}}+3 \times 4^{\mathrm{n}+2}+\mathrm{k}$ is divisible by $9, \forall \mathbf{n} \in \mathbf{N}$
$P(1)=10^1+3 \times 4^{1+2}+k=10+3 \times 64+k$
$=10+192+k=202+k$ 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 $k=5$ View full question & answer→MCQ 1541 Mark
For each $n \in N, 10^{2 n-1}+1$ is divisible by
View full question & answer→MCQ 1551 Mark
For any natural number $n, 2^{2n} - 1$ is divisible by:
Answer
- 3
Solution:
Let $P(n)=2^{2 n}-1$
Substituting $n=1,2,3, \ldots$.
$P(1)=2^{2(1)}-1=4-1=3$
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.. $P(k): 2^{2 k}-1$ is divisible by 3 , i.e., $2^{2 k}-1=3 q$, where $q \in N$
Now,
$P(k+1): 2^{2(k+1)}-1$
$=2^{2 k+2}-1$
$=2^{2 k} \times 2^2-1$
$=2^{2 k} \times 4-1$
$=3.2^{2 k}+\left(2^{2 k}-1\right)$
$=3.2^{2 k}+3 q$
$=3\left(2^{2 k}+q\right)=3 m, \text { where } m \in N$
Thus $P(k+1)$ is true, whenever $P(k)$ is true.
Therefore, for any natural number $n, 2^{2 n}-1$ is divisible by 3 View full question & answer→MCQ 1561 Mark
$2^{3 n}-7 n-1$ is divisible by:
View full question & answer→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.
AnswerCorrect 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)
$=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{\text{k}}}>\sqrt{\text{k}}\text{ is true}$
$\therefore\ \text{P (k + 1)}=\ \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{\text{k}}}+\frac{1}{\sqrt{\text{k + 1}}}>\sqrt{\text{k+1}}\text{ has to be true}.$
$\text{L.H.S}>\sqrt{\text{k}}+\frac{1}{\sqrt{\text{k + 1}}}=\frac{\sqrt{\text{k}\ (\text{k + 1)}+1}}{\sqrt{\ \text{k + 1}}}$
$\text{since}\ \sqrt{\text{k}\ (\text{k} + 1)}>\text{k}\ (\forall\text{k}\geq0)$
$\therefore\frac{\sqrt{\text{k}\ (\text{k + 1)}+1}}{\sqrt{\ \text{k + 1}}}>\frac{\text{k + 1}}{\sqrt{\text{k + 1}}}=\sqrt{\text{k + 1}}$
$\text{Let}\ \text{P}\ (\text{n})=\sqrt{\text{n}\ (\text{n + 1})}<(\text{k + 1})$
Statement-1 is correct.
$\text{P}\ (2)=\sqrt{2\times3}<3$
$\text{If}\ \text{P}\ \text{(k)}=\sqrt{\text{k (k + 1)}}<\text{(k + 1)}\ \text{is true}$
$\text{Now}\ \text{P}\ \text{(k + 1)}=\sqrt{\text{(k + 1)}(\text{k + 2})}<\text{(k + 2)}\ \text{has to be true}$
$\text{since}\ \text{(k + 1)}<\text{k + 2}$
$\therefore\sqrt{\text{(k + 1)}\text{(k + 2)}}<\text{(k + 2)}$
Hence Statement-2 is not correct explanation of Statement-1.
View full question & answer→MCQ 1581 Mark
For all $n \in N, 3 n^5+5 n^3+7 n$ is divisible by:
Answer
- 15
Solution:
Given number $=3 n^5+5 n^2+7 n$
Let $\mathrm{n}=1,2,3,4, \ldots \ldots \ldots$
$3 n^5+5 n^3+7 n=3 \times 1^2+5 \times 1^3+7 \times 1=3+5+7=15$
$3 n^5+5 n^3+7 n=3 \times 2^5+5 \times 2^3+7 \times 2=3 \times 32+5 \times 8+7 \times 2=96+40+14=150=15 \times 10$
$3 n^5+5 n^3+7 n=3 \times 3^5+5 \times 3^3+7 \times 3=3 \times 243+5 \times 27+7 \times 3=729+135+21=885=15 \times 59$
Since, all these numbers are divisible by 15 for $n=1,2,3, \ldots .$.
So, the given number is divisible by 15 . View full question & answer→MCQ 1591 Mark
$2^{3 n}-7 n-1$ is divisible by:
View full question & answer→MCQ 1601 Mark
The remainder when $5^{99}$ is divided by $13$, is:
View full question & answer→MCQ 1611 Mark
For each $n \mathrm{~N} \in, 3^{2 n}-1$ is divisible by:
View full question & answer→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:
Answer
- -1
Solution:
$(49)^n+16 n-1$
$\Rightarrow(1+48)^n+16 n-1$
$\Rightarrow 1+48 n+\ldots 48^n+16 n-1$
$\Rightarrow 64 n+n C_2(48)^2+n C_3(48)^3+\ldots+(48)^n$
$\Rightarrow 64\left(n+n C_2(6)^2+n C_3(6)^3 48+\ldots+(6)^n 8^{n-2}\right)$
$\therefore 49^n+16 n-1 \text { is divisible by } 64$ View full question & answer→MCQ 1631 Mark
$7^{2 n}+3^{n-1} \cdot 2^{3 n-3}$ is divisible by:
Answer
- 25
Solution:
Let $\mathrm{P}(1)=7^{2 \mathrm{n}}+3^{n-1} \cdot 2^{3 n-3}$
$P(1)=50 \Rightarrow$ Divisible by 25 View full question & answer→MCQ 1641 Mark
The sum of the series 1² + 2² + 3² + ……….. n² is:
- A
$\text{n}(\text{n+1}(2\text{n+1)}$
- B
$\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{2}$
- C
$\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{3}$
- ✓
$\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{6}$
AnswerCorrect option: D. $\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{6}$
Given, series is 1² + 2² + 3² + ……….. n²
$\text{sum=}\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{6}$
View full question & answer→MCQ 1651 Mark
Let f(n) = 8n - 3n, if n is odd natural number then f(n) is divisible by:
Answer
- 5
Solution:
$f(n)=8^n-3^n$
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 $\mathrm{f}(\mathrm{n})$ is divisible by 5 for odd n . View full question & answer→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 = TConclusion:I. A > XII. P < B:
AnswerGiven 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 → True (A ≥ P = S > T = B ≥ X)
II. P < B → False (P > T = B)
Hence, only I is True.
View full question & answer→MCQ 1671 Mark
The value of $(1+3 / 1)(1+5 / 4)(1+7 / 9) \ldots(1+2 n+1 / n 2)$ is:
- A
$(n+1)$
- B
$n+1)^2$
- C
$2(n+1)^2$
- ✓
View full question & answer→MCQ 1681 Mark
If P(n) is statement such that P(3) is true. assuming P(k) is true ⇒ P(k + 1) is true for all k ≥ 3, then P(n) is true:
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
In the given question P(n) is true for n = 3
Assuming P(k) is true ⇒ P(k + 1) is true for k ≥ 3
Hence, By the
Principle of Mathematical induction P(n) is true for all n ≥ 3.
View full question & answer→MCQ 1691 Mark
$S_n$ is divisible by the multiple of:
View full question & answer→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
- D
AnswerCorrect 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
LHS = 1/(1 ∙ 2 ∙ 3) = 1/6 and RHS = {1 × (1 + 3)}/[4 × (1 + 1)(1 + 2)] = ( 1 × 4)/(4 × 2 × 3) = 1/6.
Therefore 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)}
= (k³ + 6k² + 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)
⇒ 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)}
⇒ 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 ∈ N.
View full question & answer→MCQ 1711 Mark
If $x^{n-1}$ is divisible by $x-k$, then the least positive integral value of $k$ is:
View full question & answer→MCQ 1721 Mark
(1² + 2² + …… + n²) _____ for all values of n ∈ N:
AnswerLet P(n): (1² + 2² + ….. + n²) > n³/3.
When = 1, LHS = 1² = 1 and RHS = 1³/3 = 1/3.
Since 1 > 1/3, it follows that P(1) is true.
Let P(k) be true. Then,
P(k): (1² + 2² + ….. + k² ) > k³/3 …. (i)
Now,
1² + 2² + ….. + k²
+ (k + 1)²
= {1² + 2² + ….. + k² + (k + 1)²
> k³/3 + (k + 1)³ [using (i)]
= 1/3 ∙ (k³ + 3 + (k + 1)²) = 1/3 ∙ {k² + 3k² + 6k + 3}
= 1/3[k³ + 1 + 3k(k + 1) + (3k + 2)]
= 1/3 ∙ [(k + 1)³ + (3k + 2)]
> 1/3(k + 1)³P(k + 1):
1² + 2² + ….. + k² + (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 ∈ N.
View full question & answer→MCQ 1731 Mark
For every positive integer $n, 7^n-3^n$ is divisible by:
Answer
- 4
Solution:
Let $P(n)=7^n-3^n$
Substituting $\mathrm{n}=1,2,3, \ldots$
$P(1)=7^1-3^1=7-3=4$
$P(2)=7^2-3^2=49-9=40$
$P(3)=7^{3-} 3^3=343-27=316$
Thus, for every positive integer $n, 7^n-3^n$ is divisible by 4 . View full question & answer→MCQ 1741 Mark
For all $n \in N, 3 \times 5^{2 n+1}+2^{3 n+1}$ is divisible by:
View full question & answer→MCQ 1751 Mark
What is the sum of 1 + 2 + 3 + ... n?
AnswerCorrect option: C. $\frac{\text{n}(\text{n+1)}}{2}$
- $\frac{\text{n}(\text{n+1)}}{2}$
View full question & answer→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:
View full question & answer→MCQ 1771 Mark
For all $n \geq 2, n n^2\left(n^4>1\right)$ is divisible by:
View full question & answer→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.
AnswerCorrect option: A. n can be any odd integral value.
- n can be any odd integral value.
Solution:
$n\left(n^2-1\right)=n *(n-1) *(n+1)$
If $n$ is even, then $n-1$ and $n+1$ will be odd, therefore $n\left(n^2-1\right)$ is not divisible by 4 and therefore not divisible by 24.
Hence n has to be an odd integer. View full question & answer→MCQ 1791 Mark
If P(n) = 2 + 4 + 6 +…+ 2n, n ∈ N ,P(k) = k(k + 1) + 2 ⇒ P(k + 1) = (k + 1)(k + 2) + 2 for all k ∈ N. So, we can conclude that P(n) = n (n + 1) 2 for:
MCQ 1801 Mark
For any natural number n, $7^n-2^n$ is divisible by
Answer
- 5
Solution:
Given, $7^n-2^n$
Let $\mathrm{n}=1$
$7^n-2^n=7^1-2^1=7-2=5$
which is divisible by 5
Let $\mathrm{n}=2$
$7^n-2^n=7^2-2^2=49-4=45$
which is divisible by 5
Let $\mathrm{n}=3$
$7^n-2^n=7^3-2^3=343-8=335$
which is divisible by 5
Hence, for any natural number $n, 7^n-2^n$ is divisible by 5 View full question & answer→MCQ 1811 Mark
For all positive integers $n>1,\left\{x\left(x^{n-1}-n a^{n-1}\right) a^n(n-1)\right.$ is divisible by:
- A
$(x-a)^2$
- ✓
$x-a$
- C
$2(x-a)$
- D
$x+a$
View full question & answer→MCQ 1821 Mark
If P(n) = 2 + 4 + 6 +…. + 2 n, n ∈ N, then P(k) = k(k + 1) + 2 ⇒ P(k + 1) = (k + 1) (k + 2) + 2for all k ∈ N. So, we can conclude that P(n) = n(n + 1) + 2 for:
MCQ 1831 Mark
If $x^n- 1$ is divisible by x - k, then the least positive integral value of k is:
Answer
- 1
Solution:
Given,
$P(n): x^n-1$ is divisible by $x-k$
Let us substitute $\mathrm{n}=1,2,3$, ..
$\Rightarrow P(1): x-1$
$\Rightarrow P(2): x^2-1=(x-1)(x+1)$
$\Rightarrow P(3): x^3-1=(x-1)\left(x^2+x+1\right)$
$\Rightarrow P(4): x^4-1=\left(x^2-1\right)\left(x^2+1\right)=(x-1)(x+1)\left(x^2+1\right)$
Therefore, the least positive integral value of $k$ is 1 . View full question & answer→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 $n \hat{l} N$
- C
for $n>2$
- ✓
View full question & answer→MCQ 1851 Mark
For all $n \in N, 41^n-14^n$ is a multiple of:
View full question & answer→