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 $\map P n$ be the proposition:
 * $\ds \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$

Let the of this equation be denoted $\tuple {r, s, t, n}$.

Let $n = 0$.

Then:

Thus $\map P 0$ holds.

Let $n < 0$.

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

Then:

Thus $\map P n$ holds for all $n < 0$.

It remains to demonstrate that $\map P n$ 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 $\map P m$ be the proposition:
 * $\ds \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {\paren {n - r + n t + m} - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + \paren {n - r + n t + m} - t n} n$

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

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 $\map P j$ is true, for all $j$ such that $0 \le j \le m$, then it logically follows that $\map P {m + 1}$ is true.

This is the induction hypothesis:


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

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

Lemma
First a lemma:

Induction Step
This is the induction step:

We have that $\tuple {r, n - 1 - r + n t, t, n}$ holds.

We have also determined that if:
 * $\tuple {r, s, t, n}$ holds

and:
 * $\tuple {r, s - t, t, n - 1}$ holds

then:
 * $\tuple {r, s + 1, t, n}$ holds.

So $\map P m \implies \map P {m + 1}$.

Thus $\tuple {r, s, t, n}$ is shown to hold for infinitely many $s$.

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

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

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