<s>[INST] 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?
 [/INST] This might be irrelevant but it is equal to $2^{6n-1}+5 \cdot 9^n=2^{6n-6} \cdot 2^5+5\cdot 9^n$ 
Now since $2^6=64 \equiv -2 (mod 11)$, and $9^n \equiv -2(mod 11)$, the whole expression is equivalent to $(-2)^{n-1} \cdot 32+5 \cdot (-2)^n (mod 11) \equiv (-2)^n \cdot -16+5 \cdot (-2)^n (mod 11) \equiv 11 \cdot (-2)^n (mod 11)$, which is divisible by 11.  
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