I am trying to proof by induction:
$2*7^n\equiv 2^n(2+5n)(\mod25)$
However I can't proof the induction step:
$2*7^{(n+1)}\equiv 2^{(n+1)}(2+5(n+1))(\mod25)$
This is what I got so far:
$2*7^{(n+1)} = \\ 2*7^n*7 = (base) \\ 2*7*2^n(2+5n) = \\ 7*2^{(n+1)}(2+5n) = \\7*(2^{(n+2)}+2^{(n+1)}5n) = \\2^{(n+2)}(5+2)+2^{(n+1)}35n \equiv \\2^{(n+2)}5+2^{(n+3)}+2^{(n+1)}10n = \\ 2^{(n+2)}5+2^{(n+3)}+2^{(n+2)}5n = \\ 2^{(n+2)}(2+ 5(n+1)) (\mod 25) $
but: $\neq2^{(n+1)}(2+5(n+1))(\mod25) $
I feel like I'm missing something obvious...
Thanks in advance!
