How to find $\sum_{k=1}^n k\binom{n}{k}?$

Question

How to find the sum of series, $$\sum_{k=1}^n k\binom{n}{k}?$$

I try solve using this fact $$\sum_{k=1}^n \binom{n}{k} = 2^n$$ but it didn't help. And now i haven't some ideas. Maybe i can calculate $$\sum_{k=1}^n k\binom{n}{k}?$$ using $$\sum_{k=1}^{n-1} k\binom{n}{k}?$$ using this fact $$\binom{n}{k} = \binom{n - 1}{k - 1} + \binom{n - 1}{k}$$ I failed use it

Answer 1

Combinatorial argument

To make a basket team, you need players, and among these players, you need one captain. So, $k\binom{n}{k}$ is the number of team with $k$ players you can make among $n$ people. Therefore, $\displaystyle\sum_{k=1}^nk\binom{n}{k}$ is the number of team you can make with $n$ people. But to make a team, you can also take one person among the $n$ people, and say that it's the captain. You have $n$ possibilities to choose it. After, you can decide if each of the $n-1$ other people are in the team or not. So at the end, you can make $n2^{n-1}$ teams. Therefore $$\sum_{k=1}^nk\binom{n}{k}=n2^{n-1}.$$

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Using Binomial formula

$$\sum_{k=0}^n\binom{n}{k}x^k=(1+x)^n.$$ Therefore, \begin{align*} \sum_{k=1}^nk\binom{n}{k}&=\left.\frac{\mathrm d }{\mathrm d x}\right|_{x=1}\sum_{k=0}^n\binom{n}{k}x^k\\ &=\left.\frac{\mathrm d }{\mathrm d x}\right|_{x=1}(1+x)^n\\ &=n2^{n-1}. \end{align*}