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

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

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This is simply $ (1-n)^{2d} $ by the Binomial theorem. – Ragib Zaman Feb 4 '12 at 11:41
You're completely right. – Ali Feb 4 '12 at 11:54
Tag algebraic geometry? abelian varieties? – GEdgar Feb 4 '12 at 14:36
up vote 5 down vote accepted

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

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