..Prove by induction that
$\displaystyle 2( 4^{2n+1}) + 3^{3n+1}
$
is divisible by 11 $\displaystyle \forall n \! \in \! \mathbb{N}$
Ok, let's assume it holds for n, we'll now show it does so for n+1
$\displaystyle a_{n}=2\cdot{4^{2n+1}}+3^{3n+1}$ is divisible by 11
$\displaystyle a_{n+1}=2\cdot{4^{2n+3}}+3^{3n+4}$
$\displaystyle a_{n}-2\cdot{4^{2n+1}}=3^{3n+1}$ then $\displaystyle 3^{3}\cdot(a_{n}-2\cdot{4^{2n+1}})=3^{3n+4}$
So: $\displaystyle a_{n+1}=2\cdot{4^{2n+3}}+27\cdot{a_{n}}-54\cdot{4^{2n+1}}$
Since $\displaystyle 2\cdot{4^{2n+3}}-54\cdot{4^{2n+1}}=2\cdot{4^{2n+1}}\cdot{(16-27)}=-2\cdot{4^{2n+1}}\cdot{11}$ (divisible by 11)
and $\displaystyle a_{n}=2\cdot{4^{2n+1}}+3^{3n+1}$ is divisible by 11
It follows that $\displaystyle a_{n+1}=2\cdot{4^{2n+3}}+3^{3n+4}$ is also divisible by 11