Finding generating function for $ h_{n} = h_{n-1} + \binom{n+1}{3} + n$ Let $h_{n}$ denote the number of regions into which a convex polygonal region
with $n + 2$ sides is divided by its diagonals, assuming no three diagonals have a common point. With this is the initial condition  $h_{0} = 0$. 
$$ h_{n} = h_{n-1} + \binom{n+1}{3} + n$$, with $(n \ge 1)$
How does one find the generating function and obtain a formula for $h_{n}$?.  
The hardest thing for me is the $\binom{n+1}{3}$ term.  I tried expanding it:
$$\binom{n+1}{3} = \frac{(n+1)!}{3!(n+1-3)!} = \frac{n(n+1)(n-1)(n-2)!}{6 (n-2)!} = \frac{n(n+1)(n-1)}{6}$$ 
And that led to nowhere...Please help!
 A: Hint:
$$
\left( x^{n+1} \right)''' = (n+1)\,n\,(n-1)\;x^{n-2}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\verts{z} < 1}$:

\begin{align}
\sum_{n = 1}^{\infty}h_{n}z^{n} & =
\sum_{n = 1}^{\infty}h_{n - 1}\,z^{n} +
\sum_{n = 1}^{\infty}{n + 1 \choose 3}z^{n} +
\sum_{n = 1}^{\infty}nz^{n}
\\[5mm] \implies 
-h_{0}\ +\
\underbrace{\sum_{n = 0}^{\infty}h_{n}z^{n}}_{\ds{\equiv\ \mc{H}\pars{z}}} & =
\sum_{n = 0}^{\infty}h_{n}\,z^{n + 1} +
\sum_{n = 0}^{\infty}{n + 2 \choose 3}z^{n + 1} +
z\,\totald{}{z}\underbrace{\sum_{n = 1}^{\infty}z^{n}}
_{\ds{z \over 1 - z}}
\\[5mm] \implies
-h_{0} + \mc{H}\pars{z} & = z\,\mc{H}\pars{z} +
z\sum_{n = 0}^{\infty}{n + 2 \choose 3}z^{n} +
{z \over \pars{1 - z}^{2}}
\\[5mm] \implies
\mc{H}\pars{z} & = {h_{0} \over 1 - z} + 
{z \over 1 - z}\sum_{n = 0}^{\infty}{n + 2 \choose 3}z^{n} +
{z \over \pars{1 - z}^{3}}
\end{align}

Note that

\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 0}^{\infty}{n + 2 \choose 3}z^{n}} =
\sum_{n = 0}^{\infty}{n + 2 \choose n - 1}z^{n} =
\sum_{n = 0}^{\infty}{-4 \choose n - 1}\pars{-1}^{n - 1}z^{n}
\\[5mm]= &\
\sum_{n = 0}^{\infty}{-4 \choose n}\pars{-1}^{n}z^{n + 1}
=
z\sum_{n = 0}^{\infty}{-4 \choose n}\pars{-z}^{n} =
z\pars{1 - z}^{-4}
\end{align}

Then,
$$
\bbx{\mc{H}\pars{z} = {h_{0} \over 1 - z} + 
{z^{2} \over \pars{1 - z}^{5}} +
{z \over \pars{1 - z}^{3}}}
$$
