Sum over k of r-kt choose k by r over r-kt by z^k

Theorem
Let $n \in \Z_{\ge 0}$ be a positive integer.

Let $\map {A_n} {x, t}$ be the polynomial of degree $n$ defined as:
 * $\map {A_n} {x, t} := \dbinom {x - n t} n \dfrac x {x - n t}$

for $x \ne n t$.

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

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

for sufficiently small $z$.

Proof
From Sum over $k$ of $\paren {-1}^k$ by $\dbinom n k$ by $\dbinom {r - k t} n$ by $\dfrac r {r - k t}$ and renaming variables:
 * $\ds \sum_j \paren {-1}^j \dbinom k j \dbinom {r - j t} k \dfrac r {r - j t} = \delta_{k 0}$

where $\delta_{k 0}$ is the Kronecker delta.

Thus:


 * $\ds \sum_{j, k} \paren {-1}^j \dbinom k j \dbinom {r - j t} k \dfrac r {r - j t} w^k = 1$

We have:

Now let:
 * $x = \dfrac 1 {1 + w}$

Thus:

Also see

 * Binomial Theorem: obtained by setting $t = 0$ in this result