I need to show that $a_n= 2^n + a_{n-2}$ for $n$ is greater than or equal to $2$. Prior to that we are told that recursively define $a_0 = 1,\, a_1 = 3, a_2 = 5,\, $ and $ a_n = 3a_{n-2} + 2a_{n-3}$ for $n$ is greater than or equal to $3$.
I tried writing out a characteristic equation and got the following: $x^3 - 3x - 2 = 0$, for which the solutions I got were $x = -1, 2$. I tried solving for constants, but I ended up getting $4\over3$ and $-1\over3$. If I could get some guidance as to how to approach the problem and whether I was on the right track, I would greatly appreciate it.
The third solution that I got seemed to be zero. I did the following math:
$a_0 = 1 = C_1 + C_3\\a_1 = 3 = -C_1 - C_2 + 2C_3\\a_2 = 5 = C_1 + 2C_2 + 4C_3$
HINT: I suggest a very different approach. You know that
$$a_n=3a_{n-2}+2a_{n-3}\;,$$
and you want to show that
$$a_n=2^n+a_{n-2}\;.\tag{0}$$
In order for this to be true, it must be the case that
$$2^n=a_n-a_{n-2}=2(a_{n-2}+a_{n-3})$$
and hence that $a_{n-2}+a_{n-3}=2^{n-1}$ for each $n\ge 3$ or, equivalently, that
$$a_n+a_{n-1}=2^{n+1}\tag{1}$$
for each $n\ge 1$. Conversely, if you knew that $(1)$ was true for all $n\ge 0$, you could easily combine it with the original recurrence to deduce $(0)$. Try to prove $(1)$ by induction on $n$. Your induction hypothesis should be that $(1)$ holds for $n=k$ and $n=k-1$, say, and you’ll use it and the original recurrence to prove that $(1)$ holds for $n=k+1$.