Combinatorial proof of sum of numbers Does anyone have any insight on showing that $\sum_{i=1}^n i = {n+1\choose 2}$, through a combinatorial argument (i.e., not an algebraic argument)?

There is a way to see this inductively. Suppose you knew this formula up till $n-1$. Then, $n + 1$ choose $2$ is equal to $n$ choose $2$ plus the number of ways to choose a pair of distinct elements from $\{1,\cdots, n + 1\}$ such that one of them is the 'new' element $n+1$. This is equal to $n$ choose $1$, i.e. $n$, and so the difference between $n+1$ choose $2$ and $n$ choose $2$ must be $n$.
