Sum over k of r Choose k by s+k Choose n by -1^r-k/Proof 1

Proof
The proof proceeds by induction on $r$.

For all $r \in \Z_{>0}$, let $\map P r$ be the proposition:
 * $\ds \sum_k \binom r k \binom {s + k} n \paren {-1}^{r - k} = \binom s {n - r}$

Basis for the Induction
$\map P 0$ is the case:

Thus $\map P 0$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $\map P m$ is true, where $m \ge 0$, then it logically follows that $\map P {m + 1}$ is true.

This is the induction hypothesis:
 * $\ds \sum_k \binom m k \binom {s + k} n \paren {-1}^{m - k} = \binom s {n - m}$

from which it is to be shown that:
 * $\ds \sum_k \binom {m + 1} k \binom {s + k} n \paren {-1}^{m + 1 - k} = \binom s {n - \paren {m + 1} }$

Induction Step
This is the induction step:

So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\ds \sum_k \binom r k \binom {s + k} n \paren {-1}^{r - k} = \binom s {n - r}$

for all $s \in \R, r \in \Z_{\ge 0}, n \in \Z$.