How to prove $\sum_{k=0}^{n-j}(-1)^k{n-j\choose k}{n+j+k\choose 2j+1+k}$ equal to zero? How to prove $$\sum_{k=0}^{n-j}(-1)^k{n-j\choose k}{n+j+k\choose 2j+1+k}$$ equal to zero? Can you give me a direction to try?
 A: 
Hint: Using the coefficient of $[z^n]$ operator to denote the coefficient of $x^n$ of a series we can write
  \begin{align*}
\sum_{k=0}^{n-j}&(-1)^k\binom{n-j}{k}\color{blue}{\binom{n+j+k}{2j+1+k}}
=\sum_{k=0}^{n-j}\binom{n-j}{k}(-1)^k\color{blue}{[z^{2j+1+k}](1+z)^{n+j+k}}
\end{align*}
  Now apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$ and factor out all terms which do not depend on the index $k$. Then apply the binomial theorem and simplify.

A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}}}
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I'll assume $\ds{n \geq j}$.

\begin{align}
&\bbox[10px,#ffd]{%
\sum_{k = 0}^{n - j}\pars{-1}^{k}{n - j \choose k}
{n + j + k \choose 2j + 1 + k}}
\\[5mm] = &\
\sum_{k = 0}^{\infty}\pars{-1}^{k}{n - j \choose k}
\bracks{{-n + j + 2 \choose 2j + 1 + k}\pars{-1}^{2j + 1 + k}}
\\[5mm] = &\
-\sum_{k = 0}^{\infty}{n - j \choose k}
{-n + j + 2 \choose 2j + 1 + k} =
-\sum_{k = 0}^{\infty}{n - j \choose k}
\bracks{z^{2j + 1 + k}}\pars{1 + z}^{-n + j + 2}
\\[5mm] = &\
-\bracks{z^{2j + 1}}\pars{1 + z}^{-n + j + 2}\sum_{k = 0}^{\infty}{n - j \choose k}\pars{1 \over z}^{k}
\\[5mm] = &\
-\bracks{z^{2j + 1}}\pars{1 + z}^{-n + j + 2}
\pars{1 + {1 \over z}}^{n - j} =
-\bracks{z^{2j + 1}}{\pars{1 + z}^{2} \over z^{n - j}}
\\[5mm] = &\
-\bracks{z^{n + j + 1}}\pars{1 + z}^{2} =
-\delta_{n + j + 1,0} - 2\delta_{n + j + 1,1} -
\delta_{n + j + 1,2}
\\[5mm] = &\
\bbx{-\delta_{n + j,-1} - 2\delta_{n + j,0} -
\delta_{n + j,1}}
\end{align}
A: Assuming that $n\ge j$ and $n+j\ne-1$,
$$
\begin{align}
\sum_{k=0}^{n-j}(-1)^k\binom{n-j}{k}\binom{n+j+k}{2j+1+k}
&=-\sum_{k=0}^{n-j}\binom{n-j}{n-j-k}\binom{j-n}{2j+1+k}\tag1\\
&=-\binom{0}{n+j+1}\tag2
\end{align}
$$
Explanation:
$(1)$: $\binom{n-j}{k}=\binom{n-j}{n-j-k}$ (symmetry of Pascal's Triangle)
$\phantom{(1)\text{:}}$ $\binom{n+j+k}{2j+1+k}=(-1)^{k+1}\binom{j-n}{2j+1+k}$ (negative binomial coefficient)
$(2)$: Vandermonde's Identity
A: Expanding on the hint by @MarkusScheuer we write
$$\sum_{k=0}^{n-j} {n-j\choose k} (-1)^k {n+j+k\choose 2j+1+k}
= \sum_{k=0}^{n-j} {n-j\choose k} (-1)^k {n+j+k\choose n-j-1}.$$
This is
$$\sum_{k=0}^{n-j} {n-j\choose k} (-1)^k
[z^{n-j-1}] (1+z)^{n+j+k}
\\ = [z^{n-j-1}] (1+z)^{n+j}
\sum_{k=0}^{n-j} {n-j\choose k} (-1)^k (1+z)^k
\\ = [z^{n-j-1}] (1+z)^{n+j} (1-(1+z))^{n-j}
= (-1)^{n-j} [z^{n-j-1}] z^{n-j} (1+z)^{n+j}.$$
This  last  one is  zero  by  inspection  since  the argument  of  the
coefficient extractor  starts at $z^{n-j}$  yet we are  extracting the
coefficient on $z^{n-j-1}.$
