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?

Solution 1:

Keep going!

$64\cdot 2^{3k-1} + 45\cdot 3^k = 9(2^{3k-1} + 5\cdot3^k) + 55\cdot2^{3k-1}$

Solution 2:

hint: You may want to reconsider the way you split the terms at the end.

Note that $64(2^{3k - 1}) + 45(3^k) = 9(2^{3k - 1} + 5(3^k)) + 55(2^{3k - 1})$

Solution 3:

Hint note that: if $k$ is a even number then also the next number $k+2$ is even

$$2^{3(k+2)-1}+5\cdot3^{k+2}=2^{3k-1+6}+5\cdot3^{k+2}=64\cdot2^{3k-1}+9\cdot5\cdot3^{k}$$$$=55\cdot2^{3k-1}+9\cdot2^{3k-1}+9\cdot5\cdot3^{k}=55\cdot2^{3k-1}+9\cdot(2^{3k-1}+5\cdot3^{k})$$