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.

If $n=1$, then 
$$
4^{n} + 10\cdot 9^{2n-2} = 4+10 = 14,
$$
divisible by $7$;
if $n \geq 1$ is an integer such that $4^{n} + 10\cdot 9^{2n-2} = 7k$ for some integer $k \geq 1$, then 
$$
4^{n+1} + 10\cdot 9^{2(n+1) - 2} = 4\cdot 4^{n} + 10\cdot 9^{2n} = 4(4^{n} + 10\cdot 9^{2n-2}) = 28k,
$$
divisible by $7$.
