Prove $\sum_{k=0}^{r}\binom{n}{k}\binom{n}{r-k}\left(-1\right)^{k}=\left(-1\right)^{\frac{r}{2}}\binom{n}{\frac{r}{2}}$ for $r$ even.

Question

How it can be shown that:

$$\sum_{k=0}^{2r}\left(-1\right)^{k}\binom{n}{k}\binom{n}{2r-k}=\left(-1\right)^{r}\binom{n}{r}$$

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$$ \begin{align} \sum_{k=0}^{2r}\left(-1\right)^{k}\binom{n}{k}\binom{n}{2r-k} &=\left(-1\right)^{n}\sum_{k=0}^{2r}\binom{n}{k}\binom{k-2r-1}{n-2r+k}\tag{1}\\ &=\left(-1\right)^{n}\sum_{k=0}^{2r}\binom{n}{k}\sum_{l}^{}\binom{k}{n-l}\binom{-2r-1}{-2r+k+l}\tag{2}\\ &=\left(-1\right)^{n}\sum_{k=0}^{2r}\sum_{l}^{}\binom{n}{n-l}\binom{l}{n-k}\binom{-2r-1}{-2r+k+l}\tag{3}\\ &=\left(-1\right)^{n}\sum_{l}^{}\binom{n}{l}\binom{l-2r-1}{l-2r+n}\tag{4}\\ &=\sum_{l}^{}\binom{n}{l}\binom{n}{2r-l}\left(-1\right)^{l}\tag{5} \end{align} $$

• $(1)$: Pascal's rule and negative binomial coefficients
• $(2)$: Converse of Vandermonde's convolution
• $(3)$: applying the identity $\binom{n}{k}\binom{k}{r}=\binom{n}{r}\binom{n-r}{n-k}$
• $(4)$: Vandermonde's convolution
• $(5)$: negative binomial coefficients

The final answer depends on the alternating sign Vandermonde convolution.

It's known that:

$$\sum_{k=0}^{r}\binom{n}{k}\binom{n}{r-k}\left(-1\right)^{k}=\left(-1\right)^{\frac{r}{2}}\binom{n}{\frac{r}{2}}\tag{I}$$

For $r$ even.

So setting $2r \mapsto r$ follows the result, but how even $\text{(I)}$ can be proved?

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Source : math.wvu.edu

Answer 1

left hand side is the coefficient of $x^{r}$ from $(1-x)^{n}(1+x)^{n}$.

right hand side is the coefficient of $x^{r}$ from $(1-x^{2})^{n}$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\iverson}[1]{\left[\left[\,{#1}\,\right]\right]} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\sum_{k = 0}^{2r}\pars{-1}^{k}{n \choose k} {n \choose 2r - k}}:\ {\LARGE ?}.\quad\mbox{Note that}\ {n \choose 2r - k}_{k\ >\ 2r} = 0\ \mbox{such that} \\[5mm] & \color{#44f}{\sum_{k = 0}^{2r}\pars{-1}^{k}{n \choose k} {n \choose 2r - k}} = \sum_{k = 0}^{\color{red}{\LARGE\infty}}\pars{-1}^{k} {n \choose k}{n \choose 2r - k} \\[5mm] = & \ \sum_{k = 0}^{\infty} {n \choose k}\pars{-1}^{k}\bracks{z^{2r - k}}\pars{1 + z}^{n} = \bracks{z^{2r}}\pars{1 + z}^{\, n}\sum_{k = 0}^{\infty} {n \choose k}\pars{-z}^{k} \\[5mm] = & \ \bracks{z^{2r}}\pars{1 + z}^{\, n}\,\bracks{1 + \pars{-z}}^{\, n} = \bracks{z^{2r}}\pars{1 - z^{2}}^{\, n} = \\[5mm] = & \ \bbx{\color{#44f}{\pars{-1}^{r}{n \choose r}}} \\ & \end{align}

Answer 3

$$S=\sum_{k=0}^{2r} (-1)^k {n \choose k} {n \choose 2r-k}$$ $$S=[x^{k+2r-k}] (1-x)^n (1+x)^n= [x^{2r}] (1-x^2)^n= (-1)^r {n \choose r}.$$ Here $[x^j]$ means the co-efficient of $x^j$ in the given expression.