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?

First, show that this is true for $n=2$:


*

*$\frac{2^{3\cdot2-1}+5\cdot3^1}{11}=7\in\mathbb{N}$


Second, assume that this is true for $n$:


*

*$\frac{2^{3n-1}+5\cdot3^n}{11}=k\in\mathbb{N}$


Third, prove that this is true for $n+2$:


*

*$\frac{2^{3(n+2)-1}+5\cdot3^{n+2}}{11}=\frac{2^{3n+5}+5\cdot3^{n+2}}{11}$

*$\frac{2^{3n+5}+5\cdot3^{n+2}}{11}=\frac{2^6\cdot2^{3n-1}+3^2\cdot5\cdot3^n}{11}$

*$\frac{2^6\cdot2^{3n-1}+3^2\cdot5\cdot3^n}{11}=\frac{64\cdot2^{3n-1}+9\cdot5\cdot3^n}{11}$

*$\frac{64\cdot2^{3n-1}+9\cdot5\cdot3^n}{11}=\frac{55\cdot2^{3n-1}+9\cdot2^{3n-1}+9\cdot5\cdot3^n}{11}$

*$\frac{55\cdot2^{3n-1}+9\cdot2^{3n-1}+9\cdot5\cdot3^n}{11}=\frac{55\cdot2^{3n-1}+9(2^{3n-1}+5\cdot3^n)}{11}$

*$\frac{55\cdot2^{3n-1}+9(2^{3n-1}+5\cdot3^n)}{11}=\frac{55\cdot2^{3n-1}+9\cdot11k}{11}$ assumption used here

*$\frac{55\cdot2^{3n-1}+9\cdot11k}{11}=\frac{11(5\cdot2^{3n-1}+9k)}{11}$

*$\frac{11(5\cdot2^{3n-1}+9k)}{11}=5\cdot2^{3n-1}+9k\in\mathbb{N}$
