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?

Hint. Let $A_k = 2^{3k-1}+5\cdot 3^k = \frac{1}{2}\cdot 8^k+5\cdot 3^k$.
Once you prove that the sequence $\{A_k\}_{k\geq 1}$ fulfills the recurrence relation
$$ A_{k+2} = 11\cdot A_{k+1} -24\cdot A_k $$
you trivially have that $A_{k+2}\equiv -2\cdot A_k\pmod{11}$ holds.
$A_2=77\equiv 0\pmod{11}$ and the conclusion is straightforward.
