I have a problem with proving this identity by counting in two ways:
$\sum _{i=0} ^{n} i(n-i+1) = \binom{n+2}{3}$
I tried to select three elements from set that contains $n+2$ elements, but I do not know how I should select elements to prove left side of this equation.
Suppose your $n+2$ elements are the numbers $0,1,2,\dots,n,n+1$. Then any $3$-element subset $\{a,b,c\}$ of $\{0,1,2,\dots,n,n+1\}$ can be written in increasing order (so $a<b<c$), and $b$ must be between $0$ and $n$. Now the index $i$ in the sum on the left will correspond to the value of $b$. How many $3$-element subsets $\{a,b,c\}$ with $a<b<c$ are there such that $b=i$? Think about the number of possible values of $a$ and the number of possible values of $c$.