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

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.

• Please type your questions using MathJax Images are not searchable, and are not accessible to people using screen readers, for example. Besides, why should anyone take the trouble to answer your question if you won't take the trouble to type it? Apr 4, 2018 at 14:59
• edited, wasn't familiar with MathJax Apr 4, 2018 at 15:05
• Thank you. It's a lot easier to read now. Apr 4, 2018 at 15:11

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