# 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.

How it can be shown that:

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

\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?

Source : math.wvu.edu

$$S=\sum_{k=0}^{2r} (-1)^k {n \choose k} {n \choose 2r-k}$$ $$S=[x^{k+2r-k}] (1-x)^n (1+x)^n= [x^{2r}] (1-x^2)^n= (-1)^r {n \choose r}.$$ Here $$[x^j]$$ means the co-efficient of $$x^j$$ in the given expression.
• So can we say $$\left[x^{i}\right]\left[x^{j}\right]\left(...\right)=\left[x^{i+j}\right]\left(...\right)$$? – user715522 Mar 16 '20 at 9:50
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 $$(1-x^{2})^{n}$$.