Induction divisibility question Q. Prove by induction that $2^{3n-1} + 5(3^n)$ is divisible by $11$ for any even number $n$, where $n$ is an element of natural numbers.
What is have so far:
(base case): $p(2) = 77$, $77/11 = 7$. so base case holds
$p(k) = 2^{3k-1} +5(3^k) $
$p(k+2) = 2^{3k+5} + 5(3^{k+2}) $
$p(k+2) = 2^{3k+1}2^4 + 5(3^{k})(3^2) $
$p(k+2) = 2^{3k+1+(1-1)}2^4 + 5(3^{k})3^2 $
$p(k+2) = 2^{3k-1}2^6 +5(3^{k})3^2$
I am new to induction and I don't know how to continue.
A point in the right direction would be greatly appreciated, thank you.

Hint  $\ {\rm mod}\ 11\!:\,\ 2x \equiv 8^n - 3^n\equiv (-3)^n-3^n\equiv 0\ $ by $\,n\,$ even.
