How to compute this combinatoric sum?

Question

I have the sum

$$\sum_{k=0}^{n} \binom{n}{k} k$$

I know the result is $n 2^{n-1}$ but I don't know how you get there. How does one even begin to simplify a sum like this that has binomial coefficients.

Answer 1

Hint:

Just differentiate $(1+x)^n = \sum_{k=0}^n\binom{n}{k}x^k$ and set $x= 1$

Answer 2

Hint:

Use the equality $$\binom{n}{k}k=n\binom{n-1}{k-1}$$

Answer 3

We have $$\sum_{k=0}^{n} k\binom{n}{k}= \sum_{k=1}^{n-1} k\binom{n}{k}+n=\sum_{k=1}^{n-1} n\binom{n-1}{k-1}+n=n\sum_{k=0}^{n-1} \binom{n-1}{k}=n2^{n-1},$$

where the last equality follows from the binomial theorem :$$2^l=(1+1)^l=\sum_{k=0}^{l} \binom{l}{k}1^k1^{l-k}=\sum_{k=0}^{l} \binom{l}{k}.$$