Prove $a_n =3\cdot2^{n-1} +2(-1)^n$ $a_1=1, a_2=8$ and $a_n=a_{n-1} +2a_{n-2}$ for $n\ge3$

Prove that 

$$a_n =3\cdot2^{n-1} +2(-1)^n$$
for $n\in N$

Base Case: n=1 

$a_1= 3\cdot2^0+2(-1)^1, a_1 =1$, the base case holds true

I.H: Suppose its true for $$a_k =3\cdot2^{k-1} +2(-1)^k$$

How do i proceed for induction step, am i suppose to use $a_n=a_{n-1} +2a_{n-2}$ in the induction step?
NOTE: For this problem, you need two base cases, since the recursive sequence is defined in terms of the previous two terms. I'll leave that part up to you.

INDUCTION STEP: Assume that both
$$a_n=3(2^{n-1})+2(-1)^n$$
and
$$a_{n+1}=3(2^{n})+2(-1)^{n+1}$$
for some $n\ge 3$. Then, by hypothesis
$$a_{n+2}=a_{n+1}+2a_n$$
so substituting the values of $\,a_{n+1}\,$ and $\,a_{n}$ from our induction hypothesis yields
$$\begin{align} a_{n+2}&amp;=3(2^{n})+2(-1)^{n+1}+2(3(2^{n-1})+2(-1)^n)\\
&amp;=3(2^{n})+2(-1)^{n+1}+3(2^{n})+4(-1)^n\\
&amp;=6(2^{n})+2(-1)^{n}\\
&amp;=3(2^{n+1})+2(-1)^{n+2}
\end{align}$$
which completes the proof of the inductive step.
hint

Let for $n\ge 3$,
$$b_n=a_n+a_{n-1} $$

then
$$b_n=2b_{n-1} $$
Geometric.