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:
- $\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$
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:
\(\ds T_n\) | \(=\) | \(\ds \frac n 2 \left({\left({3 - 2}\right) n - 3 + 4}\right)\) | Closed Form for Polygonal Numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac n 2 \left({n + 1}\right)\) |
Hence the result.
$\blacksquare$