Let $n>0$ be an integer. Let $d$ be a positive integer.
How do I show that $\sum_{j=0}^{2d} (-1)^j n^j \binom{2d}{j} = (n-1)^{2d}?$
Let $n>0$ be an integer. Let $d$ be a positive integer.
How do I show that $\sum_{j=0}^{2d} (-1)^j n^j \binom{2d}{j} = (n-1)^{2d}?$
Isn't this just the binomial expansion of $(1-n)^{2d}$?