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 the of this equation be denoted $\left({r, s, t, n}\right)$.

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 $n$.

The procedure is to substitute $n - r + n t + m$ for the variable $s$ and establish that the identity holds for all $m \ge 0$.

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

That is:
 * $\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {n - r + m + t k} {n - k} \frac r {r - t k} = \binom {n + m} n$

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:
 * $\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {n - r + m + t k} {n - k} \frac r {r - t k} = \binom {n + m} n$

from which it is to be shown that:
 * $\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {n - r + m + 1 + t k} {n - k} \frac r {r - t k} = \binom {n + m + 1} n$

Lemma
First a lemma:

Induction Step
This is the induction step:

We have that $\left({r, n - 1 - r + n t, t, n}\right)$ holds.

We have also determined that if:
 * $\left({r, s, t, n}\right)$ holds

and:
 * $\left({r, s - t, t, n - 1}\right)$ holds

then:
 * $\left({r, s + 1, t, n}\right)$ holds.

So $P \left({m}\right) \implies P \left({m + 1}\right)$.

Thus $\left({r, s, t, n}\right)$ is shown to hold for infinitely many $s$.

As both the and  are polynomials in $s$ it follows that $\left({r, s, t, n}\right)$ holds for all $s$.

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