TITLE: Find the value of $\sum_{k=1}^{n}k\binom{n}{k}$?
QUESTION [6 upvotes]: 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 $k$ ?

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\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}