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.
 A: The next step I would take would be to write this in terms of $p(k)$ somehow:
$$p(k+2)=(2^{3k-1}+5(3^k))3^2+(2^6-3^2)2^{3k-1},$$
and see what the remainder is.
A: You want to relate $p(k+2)$ back to $p(k)$, and you've gotten most of the way there (though you could've gone straight from $p(k+2)=2^{3k+5}+5(3^{k+2})$ to $p(k+2)=2^{3k-1}2^6+5(3^k)(3^2)$).
It would be nice if instead of $2^{3k-1}2^6+5(3^k)(3^2)$ you had something like $2^{3k-1}2^6+5(3^k)(2^6)$ because then you could just factor this as $2^6\cdot p(k)$, which is divisible by $11$. If we try to re-write $p(k+2)$ to get those terms we want, we get
\begin{align*}
p(k+2)&=2^{3k-1}2^6+5(3^k)(3^2)\\
&=2^{3k-1}2^6+5(3^k)(2^6)-5(3^k)(2^6)+5(3^k)(3^2)\\
&=2^6\cdot p(k) +5(3^k)(3^2-2^6).
\end{align*}
We know $p(k)$ is divisible by $11$, so this sum is divisible by $11$ if and only if the second term $5(3^k)(3^2-2^6)$ is divisible by $11$. Simplify the powers and see what you get.
A: Hint  $\ {\rm mod}\ 11\!:\,\ 2x \equiv 8^n - 3^n\equiv (-3)^n-3^n\equiv 0\ $ by $\,n\,$ even.
