-1^n by -n choose k-1 equals -1^k by -k choose n-1
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The validity of the material on this page is questionable.
Can someone explain the disappearance of a factor of $-1$? Seems to have to do with a negative bottom index
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues.
Theorem
Let $n, k \in \Z_{\ge 0}$.
Then:
- $\paren {-1}^n \dbinom {-n} {k - 1} = \paren {-1}^k \dbinom {-k} {n - 1}$
Proof
| \(\displaystyle \paren {-1}^n \dbinom {-n} {k - 1}\) | \(=\) | \(\displaystyle \paren {-1}^n \paren {\paren {-1}^{-n - \paren {k - 1} } \binom {- \paren {k - 1 + 1} } {- n - \paren {k - 1} } }\) | Moving Top Index to Bottom in Binomial Coefficient | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \paren {-1}^{- \paren {k - 1} } \binom {- k} {- \paren {n - 1} - k}\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \paren {-1}^k \binom {- k} {- k - \paren {n - 1} }\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \paren {-1}^k \binom {- k} {n - 1}\) | Symmetry Rule for Binomial Coefficients |
Can someone explain the disappearance of a factor of $-1$? Seems to have to do with a negative bottom index
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues.
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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 $16$
