TITLE: Proof of $\sum_{i=1}^ni\cdot(n-i) = \binom{n+1}{3}$
QUESTION [5 upvotes]: I would like to prove combinatorially that $\sum_{i=1}^ni\cdot(n-i) = \binom{n+1}{3}$.
Algebraically, this identity is easily proved in the following way:
$LHS = (1+2+\cdot\cdot\cdot+n-1+n)\cdot n  - (1^2+2^2+\cdot\cdot\cdot+n^2)\\=n^2(n+1)/2 - n(n+1)(2n+1)/6 = (n+1)(n-1)\cdot n/6=RHS$
However, is there any combinatorial proof for this equality?

REPLY [10 votes]: Consider the three-element subsets of $\{0,1,\ldots,n\}$. There are
$\binom{n+1}3$ of them. How many have "middle element" $i$?
If the set has middle element $i$, there are $i$ choices for the
smallest element and $n-i$ choices for the largest element, so there are
$i(n-i)$ three-element subsets of $\{0,1,\ldots,n\}$ with middle element $i$.