Compute $\sum \binom{n}{k} \binom{n+k-1}{k} (-1)^k$ How can we study a series like this : 
$\sum_{k=0}^{n} \binom{n}{k}\binom{n+k-1}{k} (-1)^k$.
I thought about consider $S(x) = \sum_{k=0}^{n} \binom{n}{k}\binom{n+k-1}{k} x^k$. But the only one I've found is hypergeometric function. It's hard to analyze.
Hope there are more combinatorical ideas for finding such series.
Any hints? Maybe generating functions?
 A: $[x^k]:f(x)$ means the coefficient of $x^k$ in the function $f(x)$. So for instance
\begin{eqnarray*}
\binom{n}{k}=[x^k]: (1+x)^n.
\end{eqnarray*}
So for your sum we have
\begin{eqnarray*}
\sum_{k=0}^n (-1)^k \binom{n}{k} \binom{n+k-1}{k} &=& [x^0]: \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{(1+x)^{n+k-1}}{x^k} \\
&=& [x^0]: (1+x)^{n-1} \left(1- \frac{(1+x)}{x} \right)^n \\
&=& [x^n]: (1+x)^{n-1} (-1)^n =\color{red}{0}. 
\end{eqnarray*}
A: Here we  have  the  Chu-Vandermonde Identity  in disguise.

We     obtain         for        $n>0$:
  \begin{align*}
\color{blue}{\sum_{k=0}^n}&\color{blue}{ \binom{n}{k}\binom{n+k-1}{k} (-1)^k}\\
&=\sum_{k=0}^{n}\binom{n}{n-k}\binom{-n}{k}\tag{1}\\
&=\binom{0}{n}\tag{2}\\
&\,\,\color{blue}{=0}
\end{align*}
and the claim follows.

Comment:


*

*In (1) we use the   binomial identities  $\binom{p}{q}=\binom{p}{p-q}$ and  $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

*In (2) we apply the Chu-Vandermonde identity.  
A: You may also use shifted Legendre polynomials. Rodrigues' formula ensures
$$P_n(2x-1)=(-1)^n\sum_{k=0}^{n}\binom{n}{k}\binom{n+k}{k}(-1)^k x^k $$
hence
$$\sum_{k=0}^{n}\binom{n}{k}\binom{n+k-1}{k}(-1)^k=\sum_{k=0}^{n}\binom{n}{k}\binom{n+k}{k}(-1)^k\frac{n}{k+n}=n(-1)^n\int_{0}^{1}P_n(2x-1)x^{n-1}\,dx $$
and the RHS is zero, since $P_n(2x-1)$ is orthogonal to any polynomial with degree less than $n$.
A: Using Vandermonde's Identity and Negative Binomial Coefficients
$$
\begin{align}
\sum_k\binom{n}{k}\binom{n+k-1}{k}(-1)^k
&=\sum_k\binom{n}{n-k}\binom{-n}{k}\\
&=\binom{0}{n}\\[9pt]
&=[n=0]
\end{align}
$$
Note that $\binom{-1}{0}=1$.

Using Finite Differences
For $n\ge1$,
$$
\sum_k\binom{n}{k}\binom{n+k-1}{k}(-1)^k
=\sum_k(-1)^k\binom{n}{k}\binom{n+k-1}{n-1}
$$
which is an order $n$ repeated difference of a degree $n-1$ polynomial, and therefore, vanishes.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[15px,#ffd]{\sum_{k = 0}^{n}{n \choose k}
{n + k - 1 \choose k}\pars{-1}^{k}}
\\[5mm] = &\
\sum_{k = 0}^{n}{n \choose k}
\bracks{{-\bracks{n + k - 1} + k - 1 \choose k}\pars{-1}^{k}}\pars{-1}^{k}
\\[5mm] = &\
\sum_{k = 0}^{n}{n \choose k}{-n \choose k}\ =\
\bbox[15px,#ffc,border:1px solid navy]{\delta_{n0}}
\end{align}
