Generating function - technical issue. Which sequence is generated by $\frac{{5x - 3{x^2}}}{{{{(1 - x)}^3}}}$?  
We know that:
$$\frac{1}{{{{(1 - x)}^3}}} = \sum\limits_{j = 0}^\infty  {\left( {\begin{array}{*{20}{c}}
   {j + 2}  \\
   2  \\
\end{array}} \right){x^j}} $$
So we have:  
$$(5x - 3{x^2}) \cdot \sum\limits_{j = 0}^\infty  {\left( {\begin{array}{*{20}{c}}
   {j + 2}  \\
   2  \\
\end{array}} \right){x^j}} $$
Now it's easy to see that for $x^k$, $a_k$ can be defined by:  
$$5\left( {\begin{array}{*{20}{c}}
   {k + 1}  \\
   2  \\
\end{array}} \right) - 3\left( {\begin{array}{*{20}{c}}
   k  \\
   2  \\
\end{array}} \right)$$
Because we "used" $x^{k-1}$ and $x^{k-2}$ there may be a problem for $k=0,1$ and I've been told I need to check it directly. Can you help me with that?

First, note that
$$\frac{{5x - 3{x^2}}}{{{{(1 - x)}^3}}} = \frac{3}{(x-1)}+\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3}$$
And,
$$\frac{3}{(x-1)} = \sum \limits_{k=0}^{\infty}(-3)x^k$$
$$\frac{1}{(x-1)^2}= \sum \limits_{k=0}^{\infty}(k+1)x^k$$
$$\frac{-2}{(x-1)^3}= \sum_{k=0}^{\infty} x^k(k+1)(k+2)$$
Add them up and you get 
$$\sum \limits_{k=1}^{\infty} k(k+4)x^{k}$$
