Another quick induction question for a recursively defined sequence (with closed form formula given)

Question

I was given:

A sequence is defined recursively by $a_0 = 1$, $a_1 = 4$, and for $n\ge2$, $a_n = 5a_{n-1} - 6a_{n-2}$.

Use induction to prove that the closed form formula for $a_n$ is $a_n = 2\cdot3^n-2^n, n\ge0$.

So far for my proof, I have the following:

Basis: $a_0=1; 1=2\cdot3^0-2^0; 1=2\cdot1-1; 1=1$ and $a_1=4; 4=2\cdot3^1-2^1; 4=2\cdot3^-2; 4=4$

Inductive Step: Assume $a_n = 2\cdot3^n-2^n$. Therefore, $a_{n+1}=5a_{n+1-1}-6a_{n+1-2};$

$a_{n+1}=5a_{n}-6a_{n-1}$

$a_{n+1}=5(2\cdot3^n-2^n)-6a_{n-1}$

What do I do next?

Answer 1

Substitute $a_{n-1}=2\cdot3^{n-1}-2^{n-1}$:

$$\begin{align} a_{n+1}&=5(2\cdot3^n-2^n)-6a_{n-1}\\ &=5(2\cdot3^n-2^n)-6(2\cdot3^{n-1}-2^{n-1})\\ &=30\cdot3^{n-1}-10\cdot2^{n-1}-12\cdot3^{n-1}+6\cdot2^{n-1}\\ \end{align}$$

Can you finish?