Q1026 (Q1026): Difference between revisions

${\displaystyle \sum _{k=0}^{n}k{\binom {n}{k}}=\sum _{k=1}^{n-1}k{\binom {n}{k}}+n=\sum _{k=1}^{n-1}n{\binom {n-1}{k-1}}+n=n\sum _{k=0}^{n-1}{\binom {n-1}{k}}=n2^{n-1}}$