Closed Form for Triangular Numbers/Proof by Arithmetic Sequence

Theorem

The closed-form expression for the $n$th triangular number is:

$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$

Proof

\(\ds \sum_{i \mathop = 0}^{m - 1} \paren {a + i d}\)

\(=\)

\(\ds m \paren {a + \frac {m - 1} 2 d}\)

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 $d = 1$

\(\ds 0 + \sum_{i \mathop = 1}^n i\)

\(=\)

\(\ds \frac {n \paren {n + 1} } 2\)

\(\ds \sum_{i \mathop = 1}^n i\)

\(=\)

\(\ds \frac {n \paren {n + 1} } 2\)

$\blacksquare$