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 $0 < x < 1$?
Thanks!
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 $0 < x < 1$?
Thanks!
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,