How it can be shown that:


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

  • $(1)$: 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:


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.

  • $\begingroup$ So can we say $$\left[x^{i}\right]\left[x^{j}\right]\left(...\right)=\left[x^{i+j}\right]\left(...\right)$$? $\endgroup$ – user715522 Mar 16 '20 at 9:50
  • $\begingroup$ Yes you got it right. $\endgroup$ – Z Ahmed Mar 16 '20 at 9:54

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


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