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?