<s>[INST] 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.
 [/INST] We have
$$9^{2n-2} = (7+2)^{2n-2} = \sum_{k=0}^{2n-2} \dbinom{2n-2}k 7^k2^{2n-2-k} = 2^{2n-2} + 7M$$
Hence, we have
\begin{align}
4^n+10 \cdot 9^{2n-2} & = 4^n + 10 \cdot (2^{2n-2}+7M) = 4^n+10\cdot 4^{n-1} + 70M = 4^{n-1}(4+10)+70M\\
& = 14(5M+4^{n-1})
\end{align}
Hence, in fact we have that $14$ divides $4^n+10 \cdot 9^{2n-2}$ for all $n \in \mathbb{Z}^+$
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