Prove by induction that an expression is divisible by 11 
Prove, by induction that $2^{3n-1}+5\cdot3^n$ is divisible by $11$ for any even number $n\in\Bbb N$.

I am rather confused by this question. This is my attempt so far:

For $n = 2$
$2^5 + 5\cdot 9 = 77$
$77/11 = 7$
We assume that there is a value $n = k$ such that $2^{3k-1} + 5\cdot 3^k$ is divisible by $11$.
We show that it is also divisible by $11$ when $n = k + 2$
$2^{3k+5} + 5\cdot 3^{k+2}$
$32\cdot 2^3k + 5\cdot 9 \cdot3^k$
$32\cdot 2^3k + 45\cdot 3^k$
$64\cdot 2^{3k-1} + 45\cdot 3^k$ (Making both polynomials the same as when $n = k$)
$(2^{3k-1} + 5\cdot 3^k) + (63\cdot 2^{3k-1} + 40\cdot 3^k)$

The first group of terms $(2^{3k-1} + 5\cdot 3^k)$ is divisible by $11$ because we have made an assumption that the term is divisible by $11$ when $n=k$. However, the second group is not divisible by $11$. Where did I go wrong?

This is the same as proving that $2^{6n-1}+5\cdot3^{2n}$ is divisible by $11$ for all $n$. The case $n=1$ is obvious.
By induction hypothesis, you can assume $2^{6n-1}+5\cdot3^{2n}=11k$, which can be written
$$
2^{6n-1}=11k-5\cdot3^{2n}
$$
Now
\begin{align}
2^{6(n+1)-1}+5\cdot3^{2(n+1)}
&=2^6\cdot2^{6n-1}+45\cdot3^{2n}\\
&=2^6(11k-5\cdot3^{2n})+45\cdot3^{2n}\\
&=11\cdot 2^6k+3^{2n}(45-5\cdot64)
\end{align}
and you're done because $45-5\cdot64=-275=-11\cdot16$.
