Sum over k of r+tk choose k by s-tk choose n-k

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Then:

$\ds \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} = \sum_{k \mathop \ge 0} \dbinom {r + s - k} {n - k} t^k$

where $\dbinom {r + t k} k$ and so on denotes a binomial coefficient.


Proof

Let $\map f {r, s, t, n}$ be the function defined as:

$\ds \map f {r, s, t, n} := \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k}$


We have:

\(\ds \) \(\) \(\ds \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k}\)
\(\ds \) \(=\) \(\ds \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} \frac {r + t k} {r + t k}\)
\(\ds \) \(=\) \(\ds \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} \frac r {r + t k} + \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} \frac {t k} {r + t k}\)



This theorem requires a proof.
In particular: in progress
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.



Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Sum_over_k_of_r%2Btk_choose_k_by_s-tk_choose_n-k&oldid=535292"