Prove that $4^{2n} + 10n -1$ is a multiple of 25 Prove that if $n$ is a positive integer then $4^{2n} + 10n - 1$  is a multiple of $25$
I see that proof by induction would be the logical thing here so I start with trying $n=1$ and it is fine. Then assume statement is true and substitute $n$ by $n+1$ so I have the following:
$4^{2(n+1)} + 10(n+1) - 1$
And I have to prove that the above is a multiple of 25. I tried simplifying it but I can't seem to get it right. Any ideas? Thanks.

Here is a proof by induction. Suppose $4^{2n}+10n-1=25k$.
$$4^{2(n+1)}+10(n+1)-1$$
$$=16\cdot 4^{2n}+10n+9$$
$$=16\cdot 4^{2n}+160n-16-150n+25$$
$$=16(4^{2n}+10n-1)-150n+25$$
$$=16(25k)-25\cdot 6n+25$$
$$=25(16k-6n+1)$$
