How can I find the closed form expression for the sum below using generating functions?
$$\sum_{k=0}^{n}(k^2 + 3k + 2)$$
EDIT:
We transform $k^2+3k+2$ into $(k+2)(k+1)$. Let $a_k=(k+2)(k+1)$ and $b_k=1$. Then
$$\sum_{k=0}^n(k+1)(k+2)=\sum_{k=0}^na_kb_{n-k}\;,$$
which is the coefficient of $x^n$ in
$$\left(\sum_{n\ge 0}a_nx^n\right)\left(\sum_{n\ge 0}b_nx^n\right)\;.$$
Because, $b_n = 1$, we get that $\sum_{n\ge 0}b_nx^n = \sum_{n\ge 0}x^n = \frac{1}{1-x}$. If we differentiate $\sum_{n\ge 0}x^n = \frac{1}{1-x}$ twice, I will get $\sum_{n\ge 0}n(n-1)x^{n-2} = -\frac{2}{(x-1)^3}$. Then my changing the index, we get $\sum_{n\ge 2}n+2(n+1)x^{n} = -\frac{2}{(x-1)^3}$.
For $\sum_{n\ge 0}a_nx^n = \sum_{n\ge 0}(n+1)(n+2)x^n$. Which is, $\frac{d^2}{dx^2}(\sum_{n\ge 0}x^{n+2})$. And?
HINT: Notice that $k^2+3k+2=(k+2)(k+1)$. Let $a_k=(k+2)(k+1)$ and $b_k=1$. Then
$$\sum_{k=0}^n(k+1)(k+2)=\sum_{k=0}^na_kb_{n-k}\;,$$
which is the coefficient of $x^n$ in
$$\left(\sum_{n\ge 0}a_nx^n\right)\left(\sum_{n\ge 0}b_nx^n\right)\;.$$
To get the generating function for $\sum_{n\ge 0}a_nx^n$ you can use the observation that $(k+2)(k+1)x^k$ is the second derivative of $x^{k+2}$. Alternatively, if you know the generating function for $\sum_{n\ge 0}\binom{n+2}2x^n$, you can use the fact that $(k+2)(k+1)=2\binom{k+2}2$.