Solve recurring sequence using a generating function I have the sequence $a_n=3a_{n-1}-3a_{n-2}+a_{n-3}$, $\forall\ n \ge 3$, with $a_0=2$, $a_1=2$, $a_2=4$ being the known terms, and I want to find a non-recursive equation for $a_n$ using a generating function.
What I have done:
$$
\begin{align}
A(x)
& = \sum_{n\ge0}{a_nx^n}
  = a_0+a_1x+a_2x^2+\sum_{n\ge 3}\left({3a_{n-1}-3a_{n-2}+a_{n-3}}\right)x^n\\
& = 2+2x+4x^2+3x\sum_{n\ge 3}{a_{n-1}x^{n-1}}
  -3x^2\sum_{n\ge 3}{a_{n-2}x^{n-2}}
  +x^3\sum_{n\ge 3}{a_{n-3}x^{n-3}}\\
& = 2+2x+4x^2+3xA(x)-3x^2A(x)+x^3A(x)\\
& = \frac{2+2x+4x^2}{1-3x+3x^2-x^3}\\
& = \frac{4x^2+2x+2}{(1-x)^3}
\end{align}
$$
As shown above, I have reached a solution for $A(x)$, but I'm not sure how to use it to find a solution for $a_n$. Any tips pointing me in the right direction would be greatly appreciated.

$A(x)=2+2x+4x^2+3x\sum_{n\geq3}a_{n-1}x^{n-1}-3x^2\sum_{n\geq3}a_{n-2}x^{n-2}+x^3\sum_{n\geq3}a_{n-3}x^{n-3}$
$\Rightarrow A(x)=2-4x+4x^2+3x\sum_{n\geq1}a_{n-1}x^{n-1}-3x^2\sum_{n\geq2}a_{n-2}x^{n-2}+x^3A(x)$
$\Rightarrow A(x)=2-4x+4x^2+3xA(x)-3x^2A(x)+x^3A(x)$
$\Rightarrow (-x^3+3x^2-3x+1)A(x)=2-4x+4x^2$
$\Rightarrow A(x)=\frac{2-4x+4x^2}{-x^3+3x^2-3x+1}=\frac{2(2x^2-2x+1)}{(1-x)^3}=\frac{4}{1-x}-\frac{4}{(1-x)^2}+\frac{2}{(1-x)^3}$
I will use $\sum_{n\geq 0}x^n=\frac{1}{1-x}$
claim: $\frac{1}{(1-x)^k}=\sum_{n\geq 0}\binom{n+k-1}{n}x^n$
So, $\frac{1}{(1-x)^k}=(\sum_{n\geq 0}x^n)^k=\sum_{n\geq0}x^n(\sum_{{i_1}+{i_2}+...+{i_k}=n}1)=\sum_{n\geq0}x^n\binom{n+k-1}{n}$
So, now you need to find $a_n$ which is the coefficient of $x^n$ in the generating function
For your problem, it is $4\cdot\binom{n+1-1}{n}-4\cdot\binom{n+2-1}{n}+2\cdot\binom{n+3-1}{n}=4-4(n+1)+\frac{2(n+1)(n+2)}{2}=n^2+3n+2-4n\\=n^2-n+2$ 
