How can I simplify this sum? $\sum_{i=0}^n{n \choose i}i$ How can I simplify this sum?
$$\sum_{i=0}^n{n \choose i}i$$
I tried to use Pascal's formulas, but I did not succeed.
 A: Consider the binomial series,
$$(x+1)^n=\sum_{k=0}^{n} {n \choose k}x^k$$
Differentiating with respect to $x$ gives (assuming $x \neq 0$ so we don't divide by $0$ at $k=0$),
$$n(x+1)^{n-1}=\sum_{k=0}^{n} {n \choose k} k x^{k-1}$$
Take $x=1$.
A: It is also helpful to keep the following identity in mind:
\begin{align*}
\binom{n}{i}=\frac{n}{i}\binom{n-1}{i-1}\tag{1}
\end{align*}

We obtain
  \begin{align*}
\sum_{i=1}^n\binom{n}{i}i&=\sum_{i=1}^n\binom{n-1}{i-1}n\tag{2}\\
&=n\sum_{i=0}^{n-1}\binom{n-1}{i}\tag{3}\\
&=n2^{n-1}
\end{align*}

Comment:


*

*In (2) we start with index $i=1$, since the first term with $i=0$ is zero and we apply the binomial identity (1).

*In (3) we shift the index $i$ by one to start from $i=0$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\mc{J} & \equiv \sum_{i = 0}^{n}{n \choose i}i =
\sum_{i = 0}^{n}{n \choose n - i}\pars{n - i} =
n\sum_{i = 0}^{n}{n \choose i} - \,\mc{J}
\implies
\,\mc{J} = {1 \over 2}\,\pars{n\,2^{n}} = \bbx{\ds{2^{n - 1}\ n}}
\end{align}
