$\sum_{i=0}^n (-1)^k \cdot C(n,k)$

Question

Use the Binomial Theorem to Show

$$\sum_{i=0}^n (-1)^k \cdot C(n,k)$$

I'm not sure where to start here . . . I know it is missing an $a^{n-k}$ and a $b^{k}$ term that maybe I should set to be 1?

Show what? This isn't a question. – Thomas Andrews Apr 21 '13 at 00:19 — Thomas Andrews, Apr 21 '13 at 00:19

score 3 · Answer 1 · answered Apr 21 '13 at 00:05

3

Yes. What is $0=(1+(-1))^n$ according to the binomial theorem?

answered Apr 21 '13 at 00:05

Pedro

3

+1. The rate at which some answers are given, I won't be surprised if an answer is posted to a question even before the question is asked. – Apr 21 '13 at 00:08
@user17762 I cannot claim I'm the fastest gun in SE, though! – Pedro Apr 21 '13 at 00:11
LOL yes you are faster than light itself. – rbtLong Apr 21 '13 at 00:12
@PeterTamaroff There are a lot of competitors (including you and me) for the fastest gun in SE! – Apr 21 '13 at 00:13
@user17762 I feel Brian will get accepted though... ¬¬ – Pedro Apr 21 '13 at 00:16

score 3 · Accepted Answer · answered Apr 21 '13 at 00:06

3

HINT: You have the $b^k$ factor: it’s $(-1)^k$. The $a^{n-k}$ is invisible, because $a=\dots$?

answered Apr 21 '13 at 00:06

Brian M. Scott

1

Thanks! i get it now. So . . the answer is just as simple as $\sum_{i=0}^n C(n,k) \cdot (-1)^k \cdot 1^{n-k}$ ? – rbtLong Apr 21 '13 at 00:14
1

@rbtLong: That’s right, together with the fact that that sum is the binomial expansion of $\big(1+(-1)\big)^n$. – Brian M. Scott Apr 21 '13 at 00:18

2 Answers2