Sum of Sequence of n Choose 2/Proof 1

Proof
We can rewrite the as:


 * $\ds \sum_{j \mathop = 0}^m \dbinom {2 + j} 2$

where $m = n - 2$.

From Rising Sum of Binomial Coefficients:


 * $\ds \sum_{j \mathop = 0}^m \binom {n + j} n = \binom {n + m + 1} {n + 1}$

The result follows by setting $n = 2$ and changing the upper index from $m$ to $n - 2$.