Closed Form for Triangular Numbers/Proof by Arithmetic Progression

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Theorem

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

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


Proof

\(\displaystyle \sum_{i \mathop = 0}^{m - 1} \left({a + i d}\right)\) \(=\) \(\displaystyle m \left({a + \frac {m - 1} 2 d}\right)\) 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 $d = 1$
\(\displaystyle 0 + \sum_{i \mathop = 1}^n i\) \(=\) \(\displaystyle \frac {n \left({n+1}\right)} 2\)
\(\displaystyle \sum_{i \mathop = 1}^n i\) \(=\) \(\displaystyle \frac {n \left({n+1}\right)} 2\)

$\blacksquare$