Sum over k of r-kt choose k by z^k/Proof 1

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Theorem

Let $n \in \Z_{\ge 0}$ be a non-negative integer.


Then:

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

where $\dbinom {r - t k} k$ denotes a binomial coefficient.


Proof

From Sum over $k$ of $\dbinom r k$ by $\dbinom {s - k t} r$ by $\paren {-1}^k$ and renaming variables:

$\ds \sum_j \paren {-1}^j \binom k j \binom {r - j t} k = t^k$

Thus:

\(\ds \sum_{j, k} \binom k j \binom {r - j t} k \paren {-1}^j\) \(=\) \(\ds \sum_{k \mathop \ge 0} t^k\) when $k < 0$ we have $\dbinom {r - j t} k = 0$
\(\ds \leadsto \ \ \) \(\ds \sum_j \paren {-1}^j \sum_k \binom k j \binom {r - j t} k\) \(=\) \(\ds \frac 1 {1 - t}\) Sum of Infinite Geometric Sequence
\(\ds \leadsto \ \ \) \(\ds \sum_j \paren {-1}^j \sum_k \binom {r - j t} j \binom {r - j t - k} {j - k}\) \(=\) \(\ds \frac 1 {1 - t}\) Product of $\dbinom r m$ with $\dbinom m k$
\(\ds \leadsto \ \ \) \(\ds \sum_j \paren {-1}^j \binom {r - j t} j \sum_k \binom {r - j t - k} {j - k}\) \(=\) \(\ds \frac 1 {1 - t}\)


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