# Closed Form for Triangular Numbers/Proof by Arithmetic Progression

The closed-form expression for the $n$th triangular number is:
$\displaystyle T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$
 $\displaystyle \sum_{i \mathop = 0}^{m - 1} \left({a + i d}\right)$ $=$ $\displaystyle m \left({a + \frac {m - 1} 2 d}\right)$ Sum of Arithmetic Progression $\displaystyle \sum_{i \mathop = 0}^n \left({a + i d}\right)$ $=$ $\displaystyle \left({n + 1}\right) \left({a + \frac n 2 d}\right)$ Let $n = m - 1$ $\displaystyle \sum_{i \mathop = 0}^n i$ $=$ $\displaystyle \left({n + 1}\right) \left({\frac n 2}\right)$ Let $a = 0$ and $d = 1$ $\displaystyle 0 + \sum_{i \mathop = 1}^n i$ $=$ $\displaystyle \frac {n \left({n+1}\right)} 2$ $\displaystyle \sum_{i \mathop = 1}^n i$ $=$ $\displaystyle \frac {n \left({n+1}\right)} 2$
$\blacksquare$