Find the value of $\sum_{k=1}^{n}k\binom{n}{k}$?

Find the value of $\sum_{k=1}^{n}k\binom{n}{k}$ ?

I know that $\sum_{k=0}^{n}\binom{n}{k}= 2^{n}$ and so, $\sum_{k=1}^{n}\binom{n}{k}= 2^{n}-1$ but how to deal with $k$ ?

From the binomial theorem, we have

$$(1+x)^n=\sum_{k=0}^n\binom{n}{k}x^k\tag 1$$

Differentiating $(1)$ reveals

$$n(1+x)^{n-1}=\sum_{k=0}^n\binom{n}{k}kx^{k-1}\tag2$$

Setting $x=1$ in $(2)$ yields

$$n2^{n-1}=\sum_{k=0}^n\binom{n}{k}k$$

And we are done!

Interestingly, I showed in THIS ANSWER, that for $m<n$, we have $$\sum_{k=0}^n\binom{n}{k}(-1)^k k^m=0$$

• Nice Method !! Thanks :) – Jon Garrick Mar 10 '17 at 5:07
• @JonGarrick You're quite welcome. My pleasure. -Mark – Mark Viola Mar 10 '17 at 5:08

There is also a combinatorial argument:

Suppose you have a room of $n$ people and want to select a committee of $k$ of them, where one member is the chairperson. There are $k\binom{n}{k}$ ways to do this. Your sum represents the total number of ways to select such a committee of any size (from $1$ to $n$) with a chairperson.

How else can we think of this? Instead, first pick the committee chairperson. There are $n$ ways to do this. Then, go to each of the remaining $n-1$ people and decide if they should be in the committee. There are $2^{n-1}$ ways to do this. Note that we can create any committee/chairperson team this way, as before. Hence your sum is equal to $n 2^{n-1}$.

• My favourite kind of answer :) – Patrick Stevens Mar 10 '17 at 19:51
• Agreed, nothing beats a combinatorial proof. – Alex B. Mar 11 '17 at 2:04

We have \begin{align*} \sum_{k=0}^nk{n\choose k} &=\sum_{k=1}^nk{n\choose k}\\ &=n\sum_{k=1}^n\frac{(n-1)!}{(k-1)!(n-k)!}\\ &=n\sum_{\ell=0}^{n-1}\frac{(n-1)!}{\ell!((n-1)-\ell)!} \tag{by taking $\ell=k-1$}\\ &=n\sum_{\ell=0}^{n-1}{n-1\choose\ell}\\ &=n2^{n-1}. \end{align*}

Proof without derivatives:

$$\sum_{k=1}^nk\binom{n}{k}=\sum_{k=1}^n k\frac{n!}{(n-k)!k!} =\sum_{k=1}^n \frac{n!}{(n-k)!(k-1)!} \\ = \sum_{k=1}^n \frac{n(n-1)!}{(n-k)!(k-1)!} = \sum_{k=1}^n n\binom{n-1}{k-1} \\ = n \sum_{k=0}^{n-1}\binom{n-1}{k}= n2^{n-1}$$

Alternate proof via probability theory:

Toss a fair coin $n$ times, find the expected no of heads. Let $N$ be the random variable denoting the number of heads. Then $E[N] = n/2$ because $N$ is the sum of $n$ bernoulli random variables with probability $1/2$. But we also know that $N$ has a binomial distribution. Hence $$E[N] =\sum_{k=1}^nk\binom{n}{k}2^{-n}$$

"And so $\sum_{k=1}^n\binom nk=2^{n-1}$". Nope: $2^n-1$.
Sketch: Notice that $$\sum_{k=1}^n k\binom nk x^{k-1}$$ is the derivative of $\sum_{k=0}^n\binom nk x^k$.
$\newcommand{\bbx}{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}{\mathcal{#1}} \newcommand{\mrm}{\mathrm{#1}} \newcommand{\pars}{\left(\,{#1}\,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mc{I} & \equiv \sum_{k = 1}^{n}k{n \choose k} = \sum_{k = 0}^{n}\pars{n - k}{n \choose n - k} = n\sum_{k = 0}^{n}{n \choose k} - \sum_{k = 0}^{n}k{n \choose k} = n\ 2^{n} - \sum_{k = 1}^{n}k{n \choose k} \\[5mm] & = 2^{n}\,n - \mc{I} \implies \bbx{\ds{\mc{I} \equiv \sum_{k = 1}^{n}k{n \choose k} = 2^{n - 1}\,n}} \end{align}