Closed Form for Triangular Numbers

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

Plainly stated: the sum of the first $n$ natural numbers is equal to $\displaystyle \frac {n \left({n+1}\right)} 2$.

This formula pops up frequently in fields as differing as calculus and computer science, and it is elegant in its simplicity.

Proof by Arithmetic Progression
Follows directly from Sum of Arithmetic Progression putting $a = 1$ and $d = 1$.

Proof by Polygonal Numbers
Triangular numbers are $k$-gonal numbers where $k = 3$.

Hence from Closed Form for Polygonal Numbers we have that $\displaystyle T_n = \frac {n \left({2 + \left({n-1}\right)\left({k-2}\right)}\right)} 2$ where $k = 3$.

The result follows after some simple algebra.