Closed Form for Triangular Numbers

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

Plainly stated: the sum of the first $n$ natural numbers is equal to $\dfrac {n \paren {n + 1} } 2$.

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

Also see

 * Compare Integral of Power for $n = 1$:
 * $\ds \int_0^b x \rd x = \frac {b^2} 2$