Closed Form for Triangular Numbers/Direct Proof

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
We have that:
 * $\displaystyle \sum_{i \mathop = 1}^n i = 1 + 2 + \cdots + n$

Consider $\displaystyle 2 \sum_{i \mathop = 1}^n i$.

Then:

So:

Historical Note
This is the method supposedly employed by Gauss who, when very young (according to the apocryphal story), calculated the sum of the numbers from $1$ to $100$ before the teacher had barely sat back down after setting the assignment.

Whether this story is actually true or not is the subject of speculation.