# Closed Form for Triangular Numbers/Proof by Arithmetic Sequence

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