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}
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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 $$ (n+2)^2=2\binom{n}{2}+5\binom{n}{1}+4\binom{n}{0} $$
robjohn
- 345,667
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The Lemma provided yields the results desired, namely: \begin{align} S_{1} &= \frac{1}{4(m-1)} \binom{2m-2}{m-2} \ S_{2} &= 2 (m^{2} + 2m -1) \binom{2m-3}{m-2} \ S_{3} &= \frac{2}{m} , (m^{3} + 8 m^{2} + 12 m - 8) \binom{2m-3}{m-2} \end{align}. – Leucippus Oct 04 '14 at 19:46
0
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 $(1+x)^m(1+x)^{m-1}=(1+x)^{2m-1}$
lab bhattacharjee
- 274,582