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

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 0:19

HINT: You have the $b^k$ factor: it’s $(-1)^k$. The $a^{n-k}$ is invisible, because $a=\dots$?
• 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 0:14
• @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 0:18
Yes. What is $0=(1+(-1))^n$ according to the binomial theorem?