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!

$2(-1)^n=-2(-1)^{n+1}\;$ and $\;4(-1)^{n-1}=4(-1)^{n+1}$.
Note your linear recurrence equation of order $2$ is the discrete version of linear second order differential equations, and can be solved likewise:
You try to obtain as solutions geometric sequences $a^n\enspace (a\ne 0)$ (the equivalent of the exponential solutions for differential equations). This is equivalent to $a$ being a solution of the quadratic equation:
$$a^2-a-2=0\qquad(\text{characteristic equation})$$
There  two integer solutions: $\;-1$ and $\;2$. Hence the general solution of the linear recurrence equation is 
$$u_n=\alpha 2^n+\beta(-1)^n,$$
the coefficients $ \alpha$ and $\beta$ being determined by the initial conditions $u_1=1$, $u_2=8$. One finds $\alpha=\frac32$ and $\beta=2$.
