How do I evaluate the sum $\sum_{j=0}^d (-1)^j n^j \binom{d}{j}$

Question

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

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

This is simply $ (1-n)^{2d} $ by the Binomial theorem. – Ragib Zaman Feb 04 '12 at 11:41 — Ragib Zaman, Feb 04 '12 at 11:41
Tag algebraic geometry? abelian varieties? – GEdgar Feb 04 '12 at 14:36 — GEdgar, Feb 04 '12 at 14:36

score 5 · Accepted Answer · answered Feb 04 '12 at 11:41

5

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

answered Feb 04 '12 at 11:41

Gerry Myerson

1 Answers1