Binomial identity of alternating sum of products of binomial coefficients taken two at a time

Question

I came across this identity a while back and wasn't able to prove it.

$$\sum_{i=1}^{n-3}\frac{\binom{n-3}{i}\binom{n+i-1}{i}}{i+1}\cdot(-1)^{i+1}= \begin{cases} 0& \text{if $n$ is odd,}\\ 2& \text{if $n$ is even.} \end{cases} $$

I could also rewrite it as $$\sum_{i=1}^{n-3}\frac{\binom{n-3}{i}\binom{n+i-1}{i+1}}{n-1}.(-1)^{i+1}$$ I tried converting it to its factorial form but it just ended up being messy. And pairing up terms (first and last, consecutive) did not seem to work at all.

Any hints?

Answer 1

Note that by Vandermonde's_identity, \begin{align*} \sum_{i=1}^{n-3}\frac{\binom{n-3}{i}\binom{n+i-1}{i}}{i+1}\cdot(-1)^{i+1}&=\frac{1}{n-1}\sum_{i=1}^{n-3}\binom{n-3}{i}\binom{n+i-1}{i+1}\cdot(-1)^{i+1}\\ &=\frac{1}{n-1}\sum_{i=1}^{n-3}\binom{n-3}{n-3-i}\binom{1-n}{i+1}\\ &=\frac{1}{n-1}\sum_{k=2}^{n-2}\binom{n-3}{n-2-k}\binom{1-n}{k}\\ &\stackrel{\text{V.I.}}{=} \frac{1}{n-1}\left(\binom{-2}{n-2}-0-(1-n)\right)\\ &=\frac{1}{n-1}\left((n-1)(-1)^n-(1-n)\right)=1+(-1)^n \end{align*} where we used the identity $\binom{-a}{b}=\binom{a+b-1}{b}(-1)^b$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[#ffd,5px]{\sum_{i = 1}^{n - 3}{{n - 3 \choose i}{n + i - 1 \choose i} \over i + 1}\,\pars{-1}^{i + 1}} = \sum_{i = 1}^{n - 3}{{n - 3 \choose i} \over i + 1}\,\pars{-1}^{i + 1}\bracks{z^{i}}\pars{1 + z}^{n + i - 1} \\[5mm] = &\ -\bracks{z^{0}}\pars{1 + z}^{n - 1}\sum_{i = 1}^{n - 3}{{n - 3 \choose i} \over i + 1}\pars{-\,{1 \over z} - 1}^{i} \\[5mm] = &\ -\bracks{z^{0}}\pars{1 + z}^{n - 1}\int_{0}^{1}\sum_{i = 1}^{n - 3}{n - 3 \choose i}\bracks{-\pars{1 + {1 \over z}}t}^{i}\,\dd t \\[5mm] = &\ -\bracks{z^{0}}\pars{1 + z}^{n - 1}\int_{0}^{1} \braces{\bracks{1 - \pars{1 + {1 \over z}}t}^{n - 3} - 1}\,\dd t \\[5mm] = &\ -\bracks{z^{0}}\pars{1 + z}^{n - 1}\bracks{% {z^{n + 1} - \pars{-1}^{n}z^{3} \over \pars{n - 2}z^{n}\pars{1 + z}} - 1} \\[5mm] = &\ 1 - {1 \over n - 2}\bracks{z^{n}}\pars{1 + z}^{n - 2} \bracks{z^{n + 1} - \pars{-1}^{n}z^{3}} \\[5mm] = &\ 1 + {\pars{-1}^{n} \over n - 2}{n - 2 \choose n - 3} = \bbx{\large 1 + \pars{-1}^{n}\,,\quad n \geq 4} \\ & \end{align}