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

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?
