Summation proof (with binomial coefficents) I am trying to prove that $\sum_{k=2}^n$ $k(k-1) {n \choose k}$=$n(n-1)2^{n-2}$.
I was initially trying to use induction, but I think a more simple proof can be done using the fact that $\sum_{k=0}^n {n \choose k}$=$2^n$.
This is how I begin to proceed:
$\sum_{k=2}^n$ $k(k-1) {n \choose k}$= $\sum_{k=0}^{n-2}$ $(k+2)(k+1) {n \choose k+2}$= $2^{n-2} *\sum_{k=0}^{n-2}$ $(k+2)(k+1)$. 
First of all, is this correct so far? And second, how would I proceed from here. 
 A: By the binomial theorem
$$\sum_{k=0}^n {n\choose k} x^k=(1+x)^n$$
Take two derivatives in $x$ and plug in $x=1$.
A: Here is a combinatorial argument. We have a group of $n$ people, and want to award one of them a gold medal, another a silver medal, and to award plastic medals to some subset of the rest (the subset may be empty). 
We count the number of ways to do this in two different ways:
1) There are $n$ ways to decide who gets the gold, and for each such way there are $n-1$ ways to decide who gets the silver. There remain $n-2$ people. We can choose a subset of these to get plastic in $2^{n-2}$ ways, for a total count of 
$$(n)(n-1)2^{n-2}.$$ 
2) For any $k\ge 2$, we can choose the $k$ people who will get some sort of medal  in $\binom{n}{k}$ ways. From these, we can choose the gold and silver medal winners in $(k)(k-1)$ ways. The rest  get plastic. That gives a total count of 
$$\sum_{k=2}^n \binom{n}{k}(k)(k-1).$$ 
A: $$k(k-1)\cdot{n \choose k}=n(n-1)\cdot{n-2 \choose k-2}$$
A: Let $a(z), b(z), c(z)$ be Exponential Generating Functions with $a(z) = \sum_{n}[(a_nz^n)/n!]$ etc.
Binomial convolution states that if $a(z) = b(z)c(z)$ then $a_n = \sum_{k=0}^n {n \choose{k}}b_kc_{n-k}$
By using the usual product of formal power series, the proof becomes quite simple
