Inductive proof of $a_n = 3*2^{n-1} + 2(-1)^n$ if $a_n = a_{n-1} + 2*a_{n-2}, a_1 = 1, a_2 = 8$ I would appreciate a little help in finalizing a proof for the following:
Let $a_n$ be the sequence defined as $a_1 = 1$, $a_2 = 8$, and $a_n = a_{n-1} + 2*a_{n-2}$ when $n \geq 3$. Prove that $a_n = 3*2^{n-1} + 2(-1)^n$.
I decided to use strong induction and show that if the statement is true for $1,...,n$ , then it is true for $n+1$ (going off the fact that the statement is true for $n=3$). I have: $$a_{n+1} = a_n + 2*a_{n-1} = 3*2^{n-1} + 2(-1)^n + 2(3*2^{n-2} + 2(-1)^{n-1}) = 3*2^n + 2(-1)^n + 4(-1)^{n-1}$$
Which is close to the result I want $(3*2^n + 2(-1)^{n+1})$ but the signs are switched for the $(-1)$ terms; therefore I suspect I missed a $-1$ somewhere, but I could not see where the error is.
I appreciate all and any help. Thank you kindly!

$$\begin{align} 2(-1)^n\!+4(-1)^{n-1} &=\, 2(-1)^{n-1}(-1\!+\!2)\\ &=\, 2(-1)^{n-1}\\ &=\, 2(-1)^{n+1}\ \text{on scaling by } (-1)^2 = 1\end{align}$$
Simpler $ $ In terms of the linear shift operator $S$ with $S a_n = a_{n+1}$ the recurrence is $\ 0 = a_{n+2}-a_{n+1} - a_n = (S^2 -S - 2)a_n = (S-2)(S+1)a_n$
Clearly  $\,S\!-\!2$ kills $2^n$ by $\,S\, 2^n = 2\cdot 2^n,\,$  and $\,S\!+\!1\,$ kills $(-1)^n,\,$ by $\,S\,(-1)^n = (-1)(-1)^n,\,$ thus their product kills $\,c\,2^n + d (-1)^n\,$ for any constants $c,d$ (using the fact that $S-2$ and $S+1$ commute since that have constant coef's, see here for details).
