A combinatorial identity $\sum_{i=0}^k \binom ni \binom{-n}{k-i} =0$ Can anyone prove the following identity for me?
$\sum_{i=0}^k
\begin{pmatrix}
n\\
i
\end{pmatrix}
\begin{pmatrix}
-n\\
k-i
\end{pmatrix}=0$ for any positive integers $n,k$.
I'm pretty sure this is correct. But I cannot convince myself with a specific proof, even though I've done a serious calculation.
It may be just a simple verification.
 A: Following the suggestion given in the first comment we will use the fact that $(1+x)^n(1+x)^{-n}=1$ together with binomial series
$$(1+x)^\alpha=\sum_{k=0}^\infty \binom{\alpha}k.$$
We get
$$1=(1+x)^n(1+x)^{-n} = \sum_{i=0}^n \binom ni x^i \sum_{j=0}^\infty \binom{-n}j x^j = \sum_{k=0}^\infty \sum_{i=0}^k \binom ni \binom{-n}{k-i} x^k.$$
Comparing the coefficients on both sides we see that
$$
\sum_{i=0}^k \binom ni \binom{-n}{k-i} =
\begin{cases} 1,& k=0 \\ 0,&\text{otherwise.}\end{cases}
$$

This is a special case of Chu=Vandermonde identity.
$$\sum_i \binom ri \binom s{k-i} = \binom{r+s}k$$
which is obtained for $r=n$ and $s=-n$.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
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 \newcommand{\ol}[1]{\overline{#1}}
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\begin{align}
\color{#f00}{\left.\sum_{j = 0}^{k}{n \choose j}{-n \choose k - j}
\right\vert_{\ k,\,n\ \in\ \mathbb{N}_{+}}} & =
\sum_{j = 0}^{k}{n \choose j}\ \overbrace{%
\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{-n} \over z^{k - j + 1}}
\,{\dd z \over 2\pi\ic}}^{\ds{{-n \choose k - j}}}
\\[3mm] & =
\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{-n} \over z^{k + 1}}\
\overbrace{\sum_{j = 0}^{k}{n \choose j}z^{j}}^{\ds{\pars{1 + z}^{k}}}
\,{\dd z \over 2\pi\ic} =
\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{k - n} \over z^{k + 1}}\
\,{\dd z \over 2\pi\ic}
\\[3mm] & = \color{#f00}{{k - n \choose k}}\tag{1}
\\[3mm] & = \color{#f00}{{n - 1 \choose k}\pars{-1}^{k}}\tag{2}
\end{align}

In any case $\pars{~\pars{1}\ \mbox{or}\ \pars{2}~}$,
$
\begin{array}{|c|}\hline\mbox{}\\
\ds{\quad\color{#f00}{\sum_{j = 0}^{k}{n \choose j}{-n \choose k - j}}} =
\ds{\color{#f00}{0}\,,\quad\forall\ k,n\ \in\ \mathbb{N}_{+}\quad}
\\ \mbox{}\\ \hline
\end{array}
$
