Sums with squares of binomial coefficients multiplied by a polynomial It has long been known that
\begin{align}
\sum_{n=0}^{m} \binom{m}{n}^{2} = \binom{2m}{m}.
\end{align}
What is being asked here are the closed forms for the binomial series
\begin{align}
S_{1} &= \sum_{n=0}^{m} \left( n^{2} - \frac{m \, n}{2} - \frac{m}{8} \right) \binom{m}{n}^{2} \\
S_{2} &= \sum_{n=0}^{m} n(n+1) \binom{m}{n}^{2} \\
S_{3} &= \sum_{n=0}^{m} (n+2)^{2} \binom{m}{n}^{2}.
\end{align}

HINT:
As $\displaystyle n\binom mn=m\binom{m-1}{n-1},$
$\displaystyle\sum_{n=0}^mn\binom mn^2=m\sum_{n=0}^m\binom mn\binom{m-1}{n-1}=m\binom{2m-1}m$ 
comparing the coefficient of $x^{2m-1}$ in $(1+x)^m(1+x)^{m-1}=(1+x)^{2m-1}$
