Proof that pre-defined $a_n$ satisfies recursion formula. (problems with understanding notation) problem
We have recursion formula for sequence which is:
$$ a_{n+2}=2a_{n+1}+3a_{n} $$
Now I want to prove that:
$$ a_n=A\cdot 3^n+B\cdot(-1)^n $$
satisfies the recursion formula. 
my approach
I would try to just insert $a_n$ to right side of the recursion formula.
$$ a_{n+2}=2\cdot(A\cdot 3^n+B\cdot(-1)^n)+3\cdot(A\cdot 3^n+B\cdot(-1)^n) $$
but I doubt this works in this way since we have $2a_{n+1}$ there is $+1$ in lower index and I don't know how to deal with lower index with arithmetic in it. I would need to first solve what is $a_{n+1}$? since I do know what $a_{n}$ but $a_{n+1}$ is unknown?
Now if someone could provide some insight on this that would be highly appreciated.
 A: When we say that $a_n = A\cdot 3^n + B\cdot (-1)^n$ then it means that $a_\color{red}{1} = A\cdot 3^\color{red}{1} + B\cdot (-1)^\color{red}{1}$, $a_\color{red}{2} = A\cdot 3^\color{red}{2} + B\cdot (-1)^\color{red}{2}$, and so on. 
Likewise,
$a_\color{red}{n+1} = A\cdot 3^\color{red}{n+1} + B\cdot (-1)^\color{red}{n+1}$ and $a_\color{red}{n+2} = A\cdot 3^\color{red}{n+2} + B\cdot (-1)^\color{red}{n+2}$. Now you have to prove that 
$$a_{n+2} = 2 a_{n+1} + 3a_n.$$
Observe that
$$2 a_{n+1} + 3a_n = 2(A\cdot 3^{n+1} + B\cdot (-1)^{n+1}) + 3(A\cdot 3^{n} + B\cdot (-1)^{n}),$$
which can be rearranged as
$$2 a_{n+1} + 3a_n = A\cdot 3^{n+1}(2+1) + B\cdot (-1)^{n+2}(-2+3).$$
That is,
$$2 a_{n+1} + 3a_n = A\cdot 3^{n+2} + B\cdot (-1)^{n+2} = a_{n+2}.$$
A: As I see it, there are two ways to approach this problem. Let me call them the forward and reverse methods. The forward method would be to recognize that the solution to the recurrence is given by characteristic roots for $f_n=af_{n-1}+bf_{n-2}$, i.e,
$$\alpha,\beta=\frac{a+\sqrt{a^2+4b}}{2}=3,-1$$
so that 
$$a_n=A\cdot 3^n+B\cdot(-1)^n.$$
The reverse method would be to substitute the proposed solution in to the recursion formula and see what happens. To the end, consider
$$a_{n+2}=2a_{n+1}+3a_{n}$$
Upon substitution for $a_n$ we get
$$A3^{n+2}+B(-1)^{n+2}=2A3^{n+1}+2B(-1)^{n+1}+3A3^{n}+3B(-1)^{n}$$
Collecting the $A$ and $B$ terms
$$3^{n+2}=2\cdot3^{n+1}+32\cdot3^{n}=3^{n+2}\quad \text{OK}$$
$$(-1)^{n+2}=2(-1)^{n+1}+3(-1)^{n}\\
(-1)^{2}=2(-1)^{1}+3=1\quad \text{OK}$$
It's all rather straightforward.
