Sum over k of r Choose k by Minus r Choose m Minus 2k

Theorem

Let $r \in \R$, $m \in \Z$.

$\ds \sum_{k \mathop \in \Z} \binom r k \binom {-r} {m - 2 k} \paren {-1}^{m + k} = \binom r m$

We have:

$\ds \paren {1 - x^2}$

$=$

$\ds \paren {1 - x} \paren {1 + x}$

$\ds \leadsto \ \ $

$\ds \paren {1 - x}^r$

$=$

$\ds \dfrac {\paren {1 - x^2}^r} {\paren {1 + x}^r}$

$\ds $

$=$

$\ds \paren {1 - x^2}^r \paren {1 + x}^{-r}$

Thus we have:

$\ds \sum_{m \mathop \in \Z} \binom r m x^m$

$=$

$\ds \paren {1 - x}^r$

$\ds $

$=$

$\ds \paren {1 - x^2}^r \paren {1 + x}^{-r}$

$\ds $

$=$

$\ds \sum_{k \mathop \in \Z} \binom r k \paren {-x^2}^k \sum_{j \mathop \in \Z} \binom {-r} j x^j$

Comparing the coefficients of $x^m$ on both sides yields the result.

$\blacksquare$