Working with binomial coefficient $\sum_{k=0}^n (-1)^k \binom nk=0$

Question

I'm sure there's an identity or some known "trick" to solve this but I couldn't find it (Looked in Wikipedia, Wolfram and here):

$$ \sum_{k=0}^n (-1)^k{n \choose k} =0 $$

I want to know how this can be proven. I tried induction but I'm not sure what to do with ${{n+1}\choose k}$.

Thanks!

Use the binomial theorem: $0=(1-1)^n=\sum_{k=0}^n(-1)^k{n\choose k}$. — carmichael561, Nov 04 '16 at 15:51
@SirJMP I understand it, it's newton's binomial theorem. but I'm not sure how to prove it... — Joe Shmoe, Nov 04 '16 at 16:04
actually, nevermind... It's no problem finding proof for the theorem... X( — Joe Shmoe, Nov 04 '16 at 16:05
Perhaps it is worth mentioning that this is true for $n>0$. (For $n=0$, the sum is equal to $0^0=1$.) — Martin Sleziak, Mar 29 '17 at 04:43

score 3 · Accepted Answer · answered Nov 04 '16 at 15:51

3

$$ (x+y)^n = \sum_{k=0}^n\left(\matrix{n\\k}\right)x^ky^{n-k} $$ set $x=-1$ and $y=1$

answered Nov 04 '16 at 15:51

Chinny84

1 Answers1