Use mathematical induction to prove that $(3^n+7^n)-2$ is divisible by 8 for all non-negative integers. Base step: $3^0 + 7^0 - 2 = 0$ and $8|0$
Suppose that $8|f(n)$, let's say $f(n)= (3^n+7^n)-2=  8k$
Then $f(n+1) = (3^{n+1}+7^{n+1})-2$
$(3*3^{n}+7*7^{n})-2$
This is the part I get stuck. Any help would be really appreciated.
Thanks.

If for $n=k, f(k)=3^k+7^k-2$
for $n=k+1,$
$\displaystyle f(k+1)-3f(k)= 3^{k+1}+7^{k+1}-2-3(3^k+7^k-2)$
$\displaystyle=7^k(7-3)+6-2=4(7^k+1)$ which is divisible by $8$ as $7^k$ is odd
So, $f(k+1)$ will be divisible by $8\iff f(k)$ is
We can start with $f(k+1)-7f(k)$ as well.
