Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 2

Theorem

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

Then:

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

$\ds \sum_k \map {A_k} {r, t} z^k = x^r$

where:

$\map {A_n} {x, t}$ is the polynomial of degree $n$ defined as:

$\map {A_n} {x, t} := \dbinom {x - n t} n \dfrac x {x - n t}$

for $x \ne n t$

$z = x^{t + 1} - x^t$.

$\ds \sum_k \dbinom {r - t k} k z^k = \frac {x^{r + 1} } {\paren {t + 1} x - t}$

Thus:

$\ds \sum_k \map {A_k} {r, t} z^k \sum_k \dbinom {s - t k} k z^k$

$=$

$\ds \frac {x^r x^{s + 1} } {\paren {t + 1} x - t}$

$\ds $

$=$

$\ds \frac {x^{\paren {r + s} + 1} } {\paren {t + 1} x - t}$

$\ds $

$=$

$\ds \sum_k \dbinom {r + s - t k} k z^k$

Equating coefficients of $z$:

$\ds \sum_k \map {A_k} {r, t} z^k \sum_k \dbinom {s - \paren {n - k} t} k z^{n - k}$

$=$

$\ds \binom {r + s - n t} n z^n$

$\ds \sum_k \dbinom {r - k t} k \dfrac r {r - k t} z^k \dbinom {s - \paren {n - k} t} k z^{n - k}$

$=$

$\ds \binom {r + s - n t} n z^n$

$\ds \sum_k \dbinom {r - k t} k \dbinom {s - \paren {n - k} t} k \dfrac r {r - k t}$

$=$

$\ds \binom {r + s - t n} n$

$\blacksquare$