How to prove $\sum_{k=0}^{n-j}(-1)^k{n-j\choose k}{n+j+k\choose 2j+1+k}$ equal to zero?

Question

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?

Answer 1

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.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}

Answer 3

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

Answer 4

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 $z^{n-j-1}.$