Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Are any of you familiar with the closed form solutions for $\sum_{k=0}^{n} k C(n,k) x^k$ and $\sum_{k=0}^{n} k^2 C(n,k) x^k$ where $0 < x < 1$?


share|cite|improve this question
The answer is yes. :-) More seriously, what do you know and what have you tried? – Did Jun 8 '11 at 16:26
Differentiate the binomial theorem. – Qiaochu Yuan Jun 8 '11 at 16:35

As Qiaochu mentions, one solution is obtained by differentiation the binomial formula. Specifically since $$(1+x)^{n}=\sum_{k=0}^{n}\binom{n}{k}x^{k}$$ differentiation with respect to $x$ yields and multiplication by $x$ tells us that $$nx(1+x)^{n-1}=\sum_{k=0}^{n}\binom{n}{k}kx^{k}.$$ Here is an alternative: Lets rearrange $$\sum_{k=0}^{n}\binom{n}{k}kx^{k}=\sum_{k=1}^{n}\frac{n!}{(k-1)!(n-k)!}x^{k}$$ Now, with the goal of recovering the binomial formula, lets pull out $nx$, because then we get $$=nx\sum_{k=1}^{n}\frac{(n-1)!}{(k-1)!(n-k)!}x^{k-1}=nx\sum_{k=0}^{n-1}\binom{n-1}{k}x^{k}=nx(1+x)^{n-1}.$$

I leave the case with $k^2$ to you, it is very similar.

Hope that helps,

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.