# 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:

 $\ds$  $\ds \sum_{j, k} \paren {-1}^j \dbinom k j \dbinom {r - j t} k \dfrac r {r - j t} w^k$ $\ds$ $=$ $\ds \sum_j \paren {-1}^j \dfrac r {r - j t} \sum_k \dbinom k j \dbinom {r - j t} k w^k$ $\ds$ $=$ $\ds \sum_j \paren {-1}^j \dfrac r {r - j t} \sum_k \dbinom {r - j t} j \dbinom {r - j t - j} {k - j} w^k$ Product of $\dbinom r m$ with $\dbinom m k$ $\ds$ $=$ $\ds \sum_j \paren {-1}^j \dfrac r {r - j t} \dbinom {r - j t} j \sum_k \dbinom {r - j t - j} {k - j} w^k$ $\ds$ $=$ $\ds \sum_j \paren {-1}^j \map {A_j} {r, t} \sum_k \dbinom {r - j t - j} {k - j} w^k$ Definition of $\map {A_j} {r, t}$ $\ds$ $=$ $\ds \sum_j \paren {-1}^j \map {A_j} {r, t} \paren {1 + w}^{r - j t - j} w^j$ Binomial Theorem

Now let:

$x = \dfrac 1 {1 + w}$
 $\ds x$ $:=$ $\ds \dfrac 1 {1 + w}$ $\ds \leadsto \ \$ $\ds z$ $=$ $\ds -\frac w {\paren {1 + w}^{1 + t} }$ $\ds \leadsto \ \$ $\ds \paren {1 + w}^{r - j k - j} w^j \paren {-1}^j$ $=$ $\ds z^r z^j$

Thus:

 $\ds \sum_j \map {A_j} {r, t} z^j \paren {1 + w}^r$ $=$ $\ds 1$ $\ds \leadsto \ \$ $\ds \sum_j \map {A_j} {r, t} z^j$ $=$ $\ds \paren {1 + w}^{- r}$ $\ds$ $=$ $\ds x^r$

$\blacksquare$