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.
Induction can certainly be used, but a more direct method will also work as follows - (The trick here is to get to 7 as far as possible, so 9 = 7+2, etc)

$4^n + 10 \times 9^{2n-2}$

= $4^n + 3 \times (7 + 2)^{2n-2}$  (mod 7)

= $4^n + 3 \times 2^{2n-2}$ (mod 7)

= $4^n + 3 \times 4^{n-1}$ (mod 7)

= $4^{n-1}(4 + 3)$ (mod 7)

= $4^{n-1}(0)$ (mod 7)

= $0$ (mod 7)

Edit:-  In fact, because the expression is even, $14$ divides it for all $n$
$4^n+10\cdot 9^{2n-2}=4\cdot 4^{n-1}+(7+3)(9^2)^{n-1}=4\cdot 4^{n-1}+(7+3)(81)^{n-1}$

$=4\cdot 4^{n-1}+7(81)^{n-1}+3(81)^{n-1}=4\cdot 4^{n-1}+7(81)^{n-1}+3(81^{n-1}-4^{n-1}+4^{n-1})$

$=4\cdot 4^{n-1}+7(81)^{n-1}+3(81^{n-1}-4^{n-1})+3\cdot 4^{n-1}=7\cdot 4^{n-1}+7(81)^{n-1}+3(81^{n-1}-4^{n-1})$

$=7\cdot 4^{n-1}+7(81)^{n-1}+3((77+4)^{n-1}-4^{n-1})$

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

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

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