Closed Form for Triangular Numbers/Proof by Recursion

Theorem
The closed-form expression for the $n$th triangular number is:
 * $\displaystyle T_n = \sum_{i \mathop = 1}^n i = \frac {n \left({n+1}\right)} 2$

Proof
Let:
 * $\displaystyle S \left({n}\right) = 1 + 2 + \cdots + n = \sum_{i \mathop = 1}^n i$

Then:

Then: