MCQ 1011 Mark
For all $\text{n Î N,} 3.52n + 1 + 23n + 1$ is divisible by:
View full question & answer→MCQ 1021 Mark
$2.4^{2\text{n+1}}+3^{3\text{n+1}}$ is divisible by: $($for all $n \in N)$
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:
View full question & answer→MCQ 1041 Mark
If $x^n- 1$ is divisible by $\text{x}-\lambda,$ then the least positive integral value of $\lambda$ is:
AnswerGiven
$x^n- 1$
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 $1.$
View full question & answer→MCQ 1051 Mark
By mathamatical induction $n(n^2− 1)$ 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 $n \in N,$ then $121n - 25n + 1900n - (-4) n$ is divisible by which one of the following?
- A
$1904$
- ✓
$2000$
- C
$2002$
- D
$2006$
AnswerCorrect 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 $2000.$
View full question & answer→MCQ 1091 Mark
If $n \in N,$ then $11^{n+2}+ 12^{2n+1}$ 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 $n \in N, 3^{2n} + 7$ is divisible by:
AnswerGiven 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 $8$
View full question & answer→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
$2^{n+3}$
AnswerCorrect option: C. $2^{n+2}$
View full question & answer→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
- D
$n^2(n^2 + 1)/3$
AnswerCorrect option: A. $n(4n^2 – 1)/3$
Let $S = 1^2 + 3^2 + 5^2 +………(2n – 1)^2$
$\Rightarrow S = {1^2 + 2^2 + 3^2 + 4^2 ………(2n – 1)^2 + (2n)^2} – {2^2 + 4^2 + 6^2 +………+ (2n)^2}$
$\Rightarrow S = {2n \times (2n + 1) \times (4n + 1)}/6 – {4n \times (n + 1) \times (2n + 1)}/6$
$\Rightarrow S = n(4n^2 – 1)/3$
View full question & answer→MCQ 1151 Mark
For each $n N \in , 3^{2n-1}$ is divisible by:
View full question & answer→MCQ 1161 Mark
For all $n \in N ,n (n+1 )(n+5 )$ is a multiple of:
View full question & answer→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
AnswerConcept:
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.
View full question & answer→MCQ 1191 Mark
If $\forall m \in N,$ then $11^{m+2}+ 12^{2m-1}$ is divisible by:
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
AnswerCorrect option: C. $ n^3+2 n^2$
View full question & answer→MCQ 1211 Mark
For all positive integral values of n, $43^{2n}- 2n + 1$ is divisible by:
View full question & answer→MCQ 1221 Mark
$n^2+ 3n$ is always divisible by which number, provided $n$ is an integer?
Answer$P(n) = n^2+ 3n$
$P(1) = 1 + 3$
$P(1) = 4$
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 $P(n)$
View full question & answer→MCQ 1231 Mark
If $x^{2n-1}+ y^{2n-1}$ is divisible by $x + y,$ if $n$ is:
View full question & answer→MCQ 1241 Mark
What will be $P(k + 1)$ for $P(n) = n^3(n + 1)?$
AnswerCorrect option: B. $ k^4+5 k^3+9 k^2+7 k+2 $
$P(n)=n^3(n+1)$
$P(k+1)=(k+1)^3(k+1+1)$
$P(k+1)=\left(k^3+3 k^2+3 k+1\right)(k+2)$
$P(k+1)=k^4+3 k^3+3 k^2+k+2 k^3+6 k^2+6 k+2$
$P(k+1)=k^4+5 k^3+9 k^2+7 k+2$
View full question & answer→MCQ 1251 Mark
For all $n \in N, 3n^5+ 5n^3+ 7n$ is divisible by:
AnswerGiven 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 $15$
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$3.5^{2n+1}+ 2^{3n+1}$ 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:
Answerd. (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.
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:
AnswerGiven 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:
- A
for $n > 1$
- ✓
for all $n \in N$
- C
for $n > 2$
- D
AnswerCorrect option: B. for all $n \in N$
View full question & answer→MCQ 1311 Mark
If $n \in N,$ then the highest positive integerwhich divides $n(n – 1)(n – 2)$ is:
View full question & answer→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:
View full question & answer→MCQ 1331 Mark
$(n^2+ n)$ is $...........$ for all $n \in N.$
AnswerConcept:
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
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:
AnswerGiven 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 $5.$
View full question & answer→MCQ 1351 Mark
If $n$ is an even number, then the digit in the units place of $2^{2n}+ 1$ will be:
AnswerSince $2^{2n}$ 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 $7$
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?
AnswerCorrect option: B. $T(k)$ is true $\Rightarrow T(k + 1)$ is true
View full question & answer→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.
Answer$5^n- 2^n$ 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 $C$
View full question & answer→MCQ 1381 Mark
The sum of the series 1 + 2 + 3 + 4 + 5 +………..n is:
AnswerCorrect option: D. $\frac{\text{n}(\text{n} + 1)}{2}$
View full question & answer→MCQ 1391 Mark
If $n$ is a positive integer, then $2. 42n + 1 + 33n + 1$ is divisible by:
View full question & answer→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
AnswerCorrect option: B. for $n ≥ 3$
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:
- A
$n > 1$
- B
$n > 2$
- C
$\forall n \in N$
- ✓
View full question & answer→MCQ 1421 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 1431 Mark
For all $n \in N, 3.5^{2n+1}+ 2^{3n+1}$ is divisible by:
AnswerLet $P(n)$ 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 $n \in N$
View full question & answer→MCQ 1441 Mark
Let $S(K) = 1+ 3 + 5…+ (2K-1) = 3+ K^2$. Then which of the following is true:
AnswerCorrect option: B. $S(K) \Rightarrow S(K +1)$
View full question & answer→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.
AnswerCorrect option: A. $N \in N.$
View full question & answer→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
$\frac{\text{n}(4\text{n}+1)}{2}$
AnswerCorrect option: B. $\frac{\text{n}(3\text{n}-1)}{2}$
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 $(n \in N)$ 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
AnswerCorrect option: A. for all $n$
View full question & answer→MCQ 1491 Mark
The smallest positive integer $n$ for which $n! <\Big(\frac{\text{n+1}}{2}\Big)^\text{n}$ holds, is:
View full question & answer→MCQ 1501 Mark
If $P(n) = 2 + 4 + 6 + .....+ 2n, n Î N ,$ then $P(k) = k(k +1) + 2 \Rightarrow P(k +1) = (k +1)(k + 2) + 2$ for all $k Î N .$ So we can conclude that $P(n) = n(n +1) + 2$ for $D:$
AnswerCorrect option: D. $\text{none}\text{ of}\text{ these}$
View full question & answer→