Sum over k of r-k Choose m by s Choose k-t by -1^k-t/Proof 1
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Theorem
Let $s \in \R, r, t, m \in \Z_{\ge 0}$.
Then:
- $\ds \sum_{k \mathop = 0}^r \binom {r - k} m \binom s {k - t} \paren {-1}^{k - t} = \binom {r - t - s} {r - t - m}$
Proof
\(\ds \sum_{k \mathop = 0}^r \binom {r - k} m \binom s {k - t} \left({-1}\right)^{k - t}\) | \(=\) | \(\ds \sum_{k \mathop = 0}^r \binom {-\left({m + 1}\right)} {r - k - m} \binom s {k - t} \left({-1}\right)^{r - t - m}\) | Moving Top Index to Bottom in Binomial Coefficient | |||||||||||
\(\ds \) | \(=\) | \(\ds \binom {s - m - 1} {r - t - m} \left({-1}\right)^{r - t - m}\) | Chu-Vandermonde Identity | |||||||||||
\(\ds \) | \(=\) | \(\ds \binom {r - t - s} {m - s}\) | Moving Top Index to Bottom in Binomial Coefficient | |||||||||||
\(\ds \) | \(=\) | \(\ds \binom {r - t - s} {r - t - m}\) | Symmetry Rule for Binomial Coefficients |
$\blacksquare$
This article, or a section of it, needs explaining. In particular: It's there in my notebook from when I did this many years ago, but I can no longer work out why I obtained the result in either line 2 or line 3 in the above. I can't make any sense of it. Mind, I used to do a lot of drugs at the time, I could have been tripping. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. 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 {{Explain}} from the code. |
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $20$