The binomial formula. how to show: $\Sigma_{k=0}^n k \binom{n}{k} = n2^{n-1}$

Question

Does anyone know how to show that: $\Sigma_{k=0}^n k \binom{n}{k} = n2^{n-1}$?

I think we are suppose to use the binomial formula for that..

Thank you!

http://math.stackexchange.com/questions/355262/closed-form-expression-for-sum-k-0n-binomnkkp-for-integers-n-p is a generalization of your question. — user37238, Apr 08 '14 at 09:07

score 2 · Answer 1 · answered Apr 08 '14 at 09:05

2

The simplest way is to take the derivative at $x=1$ of the binomial identity $$ (1+x)^n=\sum_{k=0}^n\binom{n}{k}x^k. $$

answered Apr 08 '14 at 09:05

Omran Kouba

score 1 · Answer 2 · answered Apr 08 '14 at 09:07

1

Hint: Differentiate both sides of: $(x+1)^n = \sum_{k=0}^n \binom{n}{k} \cdot x^k$ and put $x = 1$ to get the answer.

answered Apr 08 '14 at 09:07

DeepSea

2 Answers2