Show that ${49}^n+16n-1$ is divisible by 64

Let $S\left(n\right)$ be the given statement

put $n = 1$

$={49}^n+16\left(n\right)-1={49}^1+16\left(1\right)-1$

$=64$ is divisible by 64

$\therefore S\left(1\right)$ is true

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

${49}^k+16k-1=64m$ (where m is an Integer)

${49}^k=64m-16k+1$

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

$\therefore {49}^{k+1}+16(k+1)-1$

$={49}^k\cdot 49+16k+16-1$

$=(64m-16k+1)49+16k+16-1$ (from (1))

$=64m\left(49\right)-16k\left(49\right)+49+16k+15$

$=64m\left(49\right)-49\left(16k\right)+16k+64$

$=64m\left(49\right)-16k(49-1)+64$

$=64m\left(49\right)-16k\left(48\right)+64$

$=64m\left(49\right)-16k\left(4.12\right)+64$

$=64m\left(49\right)-64k\left(12\right)+64$

$=64[49m-12k+1]$ is divisible by 64

$\therefore S(k+1)$ is true

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