Show that $2.4^{2n+1}+3^{3n+1}$ is divisible by 11, $\forall n\in N$

Let $S\left(n\right)=2·4^{2n+1}+3^{3n+1}$ is divisible by 11

put $n = 1$

$=2·4^{2\left(1\right)+1}+3^{3\left(1\right)+1}=2\left(64\right)+81=209$

$=11\times \left(19\right)$ is divisible by 11

$\therefore S\left(1\right)$ is true

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

${2.4}^{(2k+1)}+3^{(3k+1)}$ is divisible by 11

$2\cdot 4^{2k+1}+3^{3k+1}=11m$                   (where $m\in z$ )

${2.4}^{2k+1}=11m-3^{3k+1}$                   .... (1)

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

$\therefore 2\cdot 4^{2(k+1)+1}+3^{3(k+1)+1}$

$=2\cdot 4^{2k+2+1}+3^{3k+3+1}$

$=2\cdot 4^{2k+1}\cdot 4^2+3^{3k+1}\cdot 3^3$

$=\left(11m-3^{3k+1}\right)\left(16\right)+3^{3k+1}\cdot 27$       (from (1))

$=11m\left(16\right)-3^{3k+1}\left(16\right)+3^{3k+1}\left(27\right)$

$=11\mathrm{m}\left(16\right)+11\left(3^{3k+1}\right)$

$11\left(16m+3^{3k+1}\right)$ is divisible by 11

$\therefore S(k+1)$ is true

By the principle of finite mathematical induction
$S\left(n\right)$ is true $\mathrm{\forall }n\in N$.