<s>[INST] 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.
 [/INST] 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,
$$2 a_{n+1} + 3a_n = A\cdot 3^{n+2} + B\cdot (-1)^{n+2} = a_{n+2}.$$
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