Show that ${3.5}^{2n+1}+2^{3n+1}$ is divisible by 17, $\mathrm{\forall }k\in N$.

Let $S\left(n\right)\mid ={3.5}^{2n+1}+2^{3n+1}$ is divisible by 17

put $n = 1$

${3.5}^{2n+1}+2^{3n+1}$

$={3.5}^{2\left(1\right)}+2^{3\left(1\right)+1}={3.5}^{2+1}+2^{3+1}={3.5}^3+2^4$

$=3\left(125\right)+16=375+16=391$

$=17\left(23\right)$ is divisible by 17

$S\left(1\right)$ is true

Assume that $S\left(k\right)$ is true for some $k\in N$

$\Rightarrow {3.5}^{2k+1}+2^{3k+1}=17m$ (where m is an Integer)

$\Rightarrow {3.5}^{2k+1}=17m-2^{3k+1}$     ..... (1)

to prove $S\left(n\right)$is true for $n = k + 1$

$\therefore {3.5}^{2(k+1)+1}+2^{3(k+1)+1}$

$={3.5}^{2k+2+1}+2^{3k+3+1}$

$=3\cdot 5^{2k+1}\cdot 5^2+2^{3k+1}\cdot 2^3$

$=\left[17m-2^{3k+1}\right]\left(25\right)+2^{3k+1}\left(8\right)$   (from (1))

$=17m\left(25\right)-17\left(2^{3k+1}\right)$

$=17\left(25m-2^{3k+1}\right)$  is divisible by 17

$\therefore S(k+1)$ is true

By the prinicple   of finite mathematical induction
$S\left(n\right)$ is true for all $n\in N$ .