How to prove $\sum_{k=1}^{n}k\binom{n}{k} = n2^{n-1}$

Question

$\sum_{k=1}^{n}k\binom{n}{k} = n2^{n-1}$

I have tried both induction and transforming both sides to get equality but no luck

I know that

$\sum_{k=1}^{n}\binom{n}{k} = 2^{n}-1$ and $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$

p.s: I couldn't find something similar of this kind anywhere. But I thinks it's something that is related to Pascals rule. Please point out little hints, thank you

Answer 1

Given: $\text{S} \ = \displaystyle\sum_{k=1}^{n}k\binom{n}{k}$

$=\displaystyle\sum_{k=1}^{n}k\times\dfrac{n}{k}\binom{n-1}{k-1}$

$=n\displaystyle\sum_{k=1}^{n}\binom{n-1}{k-1}$

$= \boxed{n \ 2^{n-1}}$

Answer 2

Hint: Develop $(1+x)^n$ with the binomial formula, and then differentiate both sides wrt $x$.

Answer 3

You can use $\binom{n}{k} = \binom{n}{n-k}$: $$ \begin{align*} 2\sum_{k=0}^n k\binom{n}{k} &= \sum_{k=0}^n k\binom{n}{k} + \sum_{k=0}^n k\binom{n}{n-k} \\ &= \sum_{k=0}^n k\binom{n}{k} + \sum_{k=0}^n (n-k)\binom{n}{k} \\ &= \sum_{k=0}^n [k+(n-k)]\binom{n}{k} \\ &= \sum_{k=0}^n n\binom{n}{k} \\ &= n2^n. \end{align*} $$ In the first line we used $\binom{n}{k} = \binom{n}{n-k}$, in the second we used the substitution $k = n-k$ in the second sum.