Sum of Sequence of Triangular Numbers/Proof 1

Proof
From Sum of Sequence of n Choose 2 we have:

and so:

But we have that:


 * $\dbinom {n + 2} 3 = \dfrac {\paren {n + 2} \paren {n + 1} n} {3 \times 2 \times 1}$

and from Binomial Coefficient with Two/Corollary:
 * $T_n = \dfrac {\paren {n + 1} n} 2 = \dbinom {n + 1} 2$

The result follows.