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 $(1-x^{2})^{n}$.