Why is $13^n\cdot 14 + 13\cdot 2^{2n}$ divisible by $9$ for any nonnegative integer $n$? Prove that statement
$13^n\cdot14+13\cdot2^{2n}$ is divisible by $9$ for any natural $n$
So, the first step is to represent $13^n$ as sum of
$(9+4)^n$ and what the next one?
Give me a hint, please.
I've no idea where to go for now.
Thanks!

One can always do it by induction.
$13^{n+1} \cdot 14 + 13\cdot 2^{2(n+1)}=$
$13^n\cdot 14\times 13 + 13\cdot 2^{2n}\times 4=$
$13^n\cdot 14\times(9+4) + 13 \cdot2^{2n}\times 4=$
$13^n\cdot 14\times 9 + 4\times (13^n\cdot 14+ 13 \cdot2^{2n})$
which is a multiple of $9$ if $13^n\cdot 14+ 13 \cdot2^{2n}$ is.
