For $n \in \mathbb{N}$, show that $4^n + 10 \times 9^{2n-2}$ is divisible by 7 For $n \in \mathbb{N}$, show that $4^n + 10 \times 9^{2n-2}$ is divisible by 7.
I'm not sure how to do this proof so any help would be appreciated.

$4^n+10\cdot 9^{2n-2}=4\cdot 4^{n-1}+(7+3)(9^2)^{n-1}=4\cdot 4^{n-1}+(7+3)(81)^{n-1}$
$=4\cdot 4^{n-1}+7(81)^{n-1}+3(81)^{n-1}=4\cdot 4^{n-1}+7(81)^{n-1}+3(81^{n-1}-4^{n-1}+4^{n-1})$
$=4\cdot 4^{n-1}+7(81)^{n-1}+3(81^{n-1}-4^{n-1})+3\cdot 4^{n-1}=7\cdot 4^{n-1}+7(81)^{n-1}+3(81^{n-1}-4^{n-1})$
$=7\cdot 4^{n-1}+7(81)^{n-1}+3((77+4)^{n-1}-4^{n-1})$
$=7\cdot 4^{n-1}+7(81)^{n-1}+3(\sum_{k=0}^{n-1}{{n-1}\choose k}77^k4^{n-1-k}-4^{n-1})$
$=7\cdot 4^{n-1}+7(81)^{n-1}+3(4^{n-1}+\sum_{k=1}^{n-1}{{n-1}\choose k}77^k4^{n-1-k}-4^{n-1})$
$=7\cdot 4^{n-1}+7(81)^{n-1}+3(\sum_{k=1}^{n-1}{{n-1}\choose k}77^k4^{n-1-k})$
