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.

$\rm\displaystyle 25\ |\ 10n\!-\!(1\!-\!4^{2n}) \iff 5\ |\ 2n - \frac{1-(-4)^{2n}}{5}.\ $ Now via $\rm\ \dfrac{1-x^k}{1-x}\, =\, 1\!+\!x\!+\cdots+x^{k-1}\ $ 
$\rm\displaystyle we\ easily\ calculate\ that, \  mod\ 5\!:\, \frac{1-(-4)^{2n}}{1-(-4)\ \ \,}\, =\, 1\!+\!1\!+\cdots + 1^{2n-1} \equiv\, 2n\ \ $ by $\rm\: -4\equiv 1$
