Closed Form for Triangular Numbers/Proof by Polygonal Numbers

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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

Triangular numbers are $k$-gonal numbers where $k = 3$.

From Closed Form for Polygonal Numbers we have that:

$P \left({k, n}\right) = \dfrac n 2 \left({\left({k - 2}\right) n - k + 4}\right)$


Hence:

\(\displaystyle T_n\) \(=\) \(\displaystyle \frac n 2 \left({\left({3 - 2}\right) n - 3 + 4}\right)\) Closed Form for Polygonal Numbers
\(\displaystyle \) \(=\) \(\displaystyle \frac n 2 \left({n + 1}\right)\)

Hence the result.

$\blacksquare$