Theorem: Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk

Let $r, s, t \in \R, n \in \Z$.
Then:
:$\ds \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$

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 {{LHS}} of this equation be denoted $\tuple {r, s, t, n}$.
Let $n = 0$.
Then:
{{begin-eqn}}
{{eqn | o = 
      | r = \tuple {r, s, t, 0}
}}
{{eqn | r = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s + t k} {- k} \frac r {r - t k}
      | c = 
}}
{{eqn | r = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s + t k} {- k} \frac r {r - t k} \delta_{k 0}
      | c = N Choose Negative Number is Zero: $\dbinom {s + t k} {- k} = 0$ for all $k > 0$
}}
{{eqn | r = \binom r 0 \binom s 0
      | c = {{Defof|Kronecker Delta}}
}}
{{eqn | r = 1
      | c = Binomial Coefficient with Zero
}}
{{eqn | r = \binom {r + s} 0
      | c = Binomial Coefficient with Zero
}}
{{eqn | r = \binom {r + s - t n} n
      | c = putting $n = 0$
}}
{{end-eqn}}
Thus $\map P 0$ holds.
Let $n < 0$.
Let $n = -m$ where $m > 0$.
Then:
{{begin-eqn}}
{{eqn | o = 
      | r = \tuple {r, s, t, n}
}}
{{eqn | r = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {-m - k} } {-m - k} \frac r {r - t k}
      | c = 
}}
{{eqn | r = 0
      | c = N Choose Negative Number is Zero: $\dbinom {s - t \paren {-m - k} } {-m - k} = 0$
}}
{{eqn | r = \binom {r + s + t m} {-m}
      | c = N Choose Negative Number is Zero: $\dbinom {r + s + t m} {-m} = 0$ for all $m$
}}
{{end-eqn}}
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 $s = n - 1 - r + n t$.
{{:Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 1/Basis for the Induction}}{{qed|lemma}}
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:
{{:Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 1/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 {{LHS}} and {{RHS}} 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 $r, s, t \in \R, n \in \Z$.
{{qed}}