Chu-Vandermonde Identity/Proof 4

Theorem

$\ds \sum_{k \mathop = 0}^n \binom r k \binom s {n - k} = \binom {r + s} n$

Let $r, s, t \in \R, n \in \Z$.

Then:

$\ds \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$

where $r, s, t \in \R, n \in \Z$.

Setting $t = 0$:

$\ds \sum_{k \mathop \ge 0} \binom r k \binom s {n - k} = \binom {r + s} n$

which is the result required.

$\blacksquare$