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 $n$
