Title says it all, really. I'm struggling to find the sum of the following equation.
$\sum_{k=2}^nk*(k-1)*{n \choose k}$
I believe I should simplify to $\sum_{k=2}^n[k^2{n \choose k}-k{n \choose k}]$, but I'm not sure where to go after that. I know from another question that $\sum_{k=0}^nk^2{n \choose k}=2^{n-1}(n^2+n)$, if that helps.
Any help is appreciated!
Two different hints:
Hint1: What happens if you take the second derivative of $(x+1)^n?$ using the binomial theorem?
Hint2: $$\binom{n}{k}\binom{k}{j}=\binom{n}{j}\binom{n-j}{k-j}.$$
Another variation is to use the binomial identity \begin{align*} k\binom{n}{k}=n\binom{n-1}{k-1} \end{align*}
We obtain \begin{align*} \sum_{k=2}^nk(k-1)\binom{n}{k}&=n\sum_{k=2}^n(k-1)\binom{n-1}{k-1}\\ &=n(n-1)\sum_{k=2}^n\binom{n-2}{k-2}\\ &=n(n-1)\sum_{k=0}^{n-2}\binom{n-2}{k}\tag{1}\\ &=n(n-1)2^{n-2} \end{align*}
In (1) we shift the index $k$ to start from $k=0$.