# Proof of $\sum_{i=1}^ni\cdot(n-i) = \binom{n+1}{3}$

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?

• What is "Combinatoric way"? – Krish Sep 14 '17 at 5:51
• @Krish OP is presumably asking for a combinatorial proof of the identity. – JMoravitz Sep 14 '17 at 5:51
• @JMoravitz thanks edited OP – Beverlie Sep 14 '17 at 5:52

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 $i$.