Question

$7^{2n}+2^{3n-3}{.3}^{n-1}$ is divisible by 25 for all $nϵ$ N.

Answer 1

Let $P\left(n\right):7^{2n}+2^{3n-3}.3^{n-1}$ is divisible by 25

For n = 1

$=7^2+2^o{.3}^o$

$=49+1$

$=50$

It is divisible of 25

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k,

$7^{2k}+2^{3k-3}{.3}^{k-1}=25\mu$ ......(1)

We have to show that,

$7^{2(k+1)}+2^{3k}{3}^k$ is divisible by 25

$7^{2(k+1)}+2^{3k}{.3}^k=25$

$7^{2(k+1)}+2^{3k}{.3}^k=25\lambda$

Now,

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

$=7^{2k}{.7}^2+2^{3k}{.3}^k$

$=(25\lambda -2^{3k-3}{.3}^{k-1})49+2^{3k}.3k$

[Using equation (1)]

$25\lambda .49-\frac{2^{3k}}{8}.\frac{3^k}{3}.49+2^{3k}{.3}^k$

$=\mathrm{24.25.49}\lambda -2^{3k}.3^k.49+{24.2}^{3k}{.3}^k$

$=\mathrm{24.25.49}\lambda -{25.2}^{3k}{.3}^k$

$=25(24.49\lambda -2^{3k}{.3}^k)$

$=25\mu$

$\Rightarrow$ P(n) is true for n = k + 1

$\Rightarrow$ P(n) is true for all $nϵN$ by PMI