Sums with squares of binomial coefficients multiplied by a polynomial

Question

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}

Answer 1

Lemma: $$ \begin{align} \sum_{n=0}^m\binom{n}{k}\binom{m}{n}^2 &=\sum_{n=0}^m\binom{n}{k}\binom{m}{n}\binom{m}{m-n}\tag{1}\\ &=\sum_{n=0}^m\binom{m}{k}\binom{m-k}{n-k}\binom{m}{m-n}\tag{2}\\ &=\binom{m}{k}\binom{2m-k}{m-k}\tag{3}\\ &=\binom{m}{k}\binom{2m-k}{m}\tag{4}\\ &=\binom{2m-k}{k}\binom{2m-2k}{m-k}\tag{5} \end{align} $$ Explanation:
$(1)$: $\binom{m\vphantom{k}}{n}=\binom{m\vphantom{k}}{m-n}$
$(2)$: $\binom{n\vphantom{k}}{k}\binom{m\vphantom{k}}{n}=\binom{m\vphantom{k}}{k}\binom{m-k}{n-k}$
$(3)$: Vandermonde's Identity$\vphantom{\binom{k}{n}}$
$(4)$: $\binom{m\vphantom{k}}{n}=\binom{m\vphantom{k}}{m-n}$
$(5)$: $\binom{n\vphantom{k}}{k}\binom{m\vphantom{k}}{n}=\binom{m\vphantom{k}}{k}\binom{m-k}{n-k}$

---

Apply the Lemma to $$ \color{#C00000}{n^2}-\frac{m}2\color{#00A000}{n}-\frac{m}8\color{#0000FF}{1}=\color{#C00000}{2\binom{n}{2}+\binom{n}{1}}-\frac{m}2\color{#00A000}{\binom{n}{1}}-\frac{m}8\color{#0000FF}{\binom{n}{0}} $$ and $$ n(n+1)=2\binom{n}{2}+2\binom{n}{1} $$ and $$ (n+2)^2=2\binom{n}{2}+5\binom{n}{1}+4\binom{n}{0} $$

Answer 2

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 $(1+x)^m(1+x)^{m-1}=(1+x)^{2m-1}$