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:

$\displaystyle \sum_k \dbinom {r - t k} k z^k = \frac {x^{r + 1} } {\left({t + 1}\right)x - t}$

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 $\left({-1}\right)^k$ and renaming variables:

$\displaystyle \sum_j \left({-1}\right)^j \binom k j \binom {r - j t} k = t^k$

Thus:

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



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