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Let $n>0$ be an integer. Let $d$ be a positive integer.

How do I show that $$\sum_{j=0}^{2d} (-1)^j n^j \binom{2d}{j} = (n-1)^{2d}?$$

  • 4
    This is simply $ (1-n)^{2d} $ by the Binomial theorem.2012-02-04
  • 0
    You're completely right.2012-02-04
  • 0
    Tag algebraic geometry? abelian varieties?2012-02-04

1 Answers 1

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Isn't this just the binomial expansion of $(1-n)^{2d}$?