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

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

Consider the equation:
 * $\displaystyle (1): \quad \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$

where $\dbinom {r - t k} k$ etc. are binomial coefficients.

Then equation $(1)$ holds for the special case where $s = n - 1 - r + n t$.

Proof
Substituting $n - 1 - r + n t$ for $s$ in the :

Substituting $n - 1 - r + n t$ for $s$ in the :

The two products give a polynomial of degree $n - 1$ in $k$.

Hence the sum for all $k$ is $0$.

Thus we have:

Thus the equation indeed holds for the special case where $s = n - 1 - r + n t$.