Combinatoric proof of $\sum_{i=1}^ni(n-i+1)={n+2\choose{3}}.$

Question

I would like to get some clue in combinatorially proving this assertion:

$$\sum_{i=1}^n i(n-i+1) =\binom{n+2}{3}.$$

In proving I mean to formulate a problem that both the right and left side answers.

Thanks.

Answer 1

To prove it algebraically, notice that:

$$\sum_{i=1}^{n}i(n-i+1)=(n+1)\sum_{i=1}^{n}i-\sum_{i=1}^{n}i^2$$

And use that: $$\sum_{i=1}^{n}i=\frac{n(n+1)}{2}\tag{1}$$ $$\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}\tag{2}$$

Answer 2

The problem is choosing $3$ elements from a set of $n+2$ elements, which the right-hand side clearly answers. To see that the left-hand side answers this question, think about the elements as lined up in a row. You pick the middle element, one on its left, and one on its right.

You take it from here.