Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 1

Proof
For all $n \in \Z$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \left({n - k}\right)} {n - k} \frac r {r - t k} = \binom {r + s - t n} n$

Let $n = 0$.

Then:

Thus $P \left({0}\right)$ holds.

Let $n < 0$.

Let $n = -m$ where $m > 0$.

Then:

Thus $P \left({n}\right)$ holds for all $n < 0$.

It remains to demonstrate that $P \left({n}\right)$ holds for all $n > 0$.

The proof continues by strong induction on $n$.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \left({n - k}\right)} {n - k} \frac r {r - t k} = \binom {r + s - t n} n$

Basis for the Induction
Consider the special case where $s = n - 1 - r + n t$.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({j}\right)$ is true, for all $j$ such that $0 \le j \le m$, then it logically follows that $P \left({m + 1}\right)$ is true.

This is the induction hypothesis:
 * $proposition_m$

from which it is to be shown that:
 * $proposition_{m + 1}$

Induction Step
This is the induction step:

So $P \left({m}\right) \implies P \left({m + 1}\right)$ and the result follows by the Second Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \left({n - k}\right)} {n - k} \frac r {r - t k} = \binom {r + s - t n} n$

for all $r, s, t \in \R, n \in \Z$.