Question

If $n\in N$ , then $7^{2n}+2^{3n-3}.3^{n-1}$ is always divisible by

• A. $25$
• B. None of these
• C. $35$
• D. $45$

Answer 1

The correct option is A $25$

$7^{2n}+2^{3n-3}\times 3^{n-1}={49}^n+\frac{2^{3n}}{2^3}\times \frac{3^n}{3}$

$\Rightarrow {49}^n+\frac{2^{3n}+3^n}{24}={49}^n+\frac{(2^3\times 3{)}^n}{24}$

$={49}^n+\frac{{24}^n}{24}$

$\Rightarrow {49}^n+{24}^{n-1}$

$49\equiv 24\left(mod25\right)$

${49}^n\equiv {24}^n\left(mod25\right)$

${49}^n\equiv 25\times m+{24}^{n}m\in$

${49}^n+{24}^{n-1}=25\times m+{24}^n+{24}^{n-1}$

${49}^n+{24}^{n-1}=25m+(24+1)\times {24}^{n-1}$

$\Rightarrow 25(m+{24}^{n-1})\equiv 0mol25\left[nϵN\right]$

So, option (A) is correct

$I{I}^{nd}$ method-

we can prove this by mathimation

induction.

for n = 1. $7^{2n}+2^{3n-3}{.3}^{n-1}$

$49+1\times 1=50$

it is divisible by 1, 2, 5, 10, 25 ,50

so, option (c), (D), us discord

Ans Now, if it is true for $\forall n$ then

option (A) is correct.

So, let the result be one true for n=k i.e

$7^{2k}+2^{3(k-1)},3^{k-1}$ is a multiple of 25

Now, we need to prove that the result is also true for $n=k+1,$ that is

$7^{2k+2}+2^{3k}{.3}^k$ is a multiple of 25

$\Rightarrow 7^{2k+2}+2^{3k}{.3}^k$

$=7^2{.7}^{2k}+2^3{.2}^{3(k-1)}.3{.3}^{k-1}$

$={49.7}^{2k}+{49.2}^{3(k-1)}{.3}^{k-1}-252^{3(k-1)}{.3}^{k-1}$

$=49(7^{2k}+2^{3(k-1)}{.3}^{k-1})-{25.2}^{3(k-1)}{.3}^{k-1}$

by owr assumption

$7^{2k}+2^{3(k-1)}{.3}^{k-1}$ is a multiple of 25.

$\Rightarrow 49(7^{2k}+2^{3(k-1){.3}^{k-1}})$ is a

multiple of 25

and also ${25.2}^{3(k-1)}{.3}^{k-1}$ is a multiple of 25

$\Rightarrow 49(7^{2k}+2^{3(k-1)}{.3}^{k-1})-{25.2}^{3-(k-1){.3}^{k-1}}$

is a multiple of 25.

$7^{2k+2}+2^{3k}{.3}^k$ is a multiple of 25.

$\therefore$ The result is true for $n=k+1$ hence,

it is true $\forall n$

So option (A) is correct.