Prove the identity combinatorially $\sum^{n-1}_{i=1} i = \frac{n(n-1)}{2}$

Question

I know the RHS is ${n}\choose 2$ and I'm suppose to think to the LHS as a pair but I'm not sure how exactly I'm suppose to decomposition it

Answer 1

You’re counting $2$-element subsets of $\{0,1,\ldots,n-1\}$. The righthand side counts them all at once. The $i$-th term on the lefthand side counts the number of pairs having $i$ as larger member.

Answer 2

Recall Pascal's triangle and the Hockey-stick identity to sum along the $1+2+3+\dots+n$ diagonal.