-1^n by -n choose k-1 equals -1^k by -k choose n-1

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Theorem

Let $n, k \in \Z_{\ge 0}$.

Then:

$\left({-1}\right)^n \dbinom {-n} {k - 1} = \left({-1}\right)^k \dbinom {-k} {n - 1}$


Proof

\(\displaystyle \left({-1}\right)^n \dbinom {-n} {k - 1}\) \(=\) \(\displaystyle \left({-1}\right)^n \left({\left({-1}\right)^{-n - \left({k - 1}\right)} \binom {- \left({k - 1 + 1}\right)} {- \left({n - \left({k - 1}\right)}\right)} }\right)\) Moving Top Index to Bottom in Binomial Coefficient
\(\displaystyle \) \(=\) \(\displaystyle \left({-1}\right)^{- \left({k - 1}\right)} \binom {- k} {- \left({n + 1}\right) - k}\)
\(\displaystyle \) \(=\) \(\displaystyle \left({-1}\right)^k \binom {- k} {- k - \left({n + 1}\right)}\)
\(\displaystyle \) \(=\) \(\displaystyle \left({-1}\right)^k \binom {- k} {n - 1}\) Symmetry Rule for Binomial Coefficients

$\blacksquare$


Sources

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