Closed Form for Triangular Numbers/Proof by Telescoping Sum
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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
Observe that:
\(\ds \sum_{i \mathop = 1}^n \paren {\paren {i + 1}^2 - i^2}\) | \(=\) | \(\ds -\sum_{i \mathop = 1}^n \paren {i^2 - \paren {i + 1} ^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {1 - \paren {n + 1}^2}\) | Telescoping Series | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {n + 1}^2 - 1\) |
Moreover, we have:
- $\paren {i + 1}^2 - i^2 = 2 i + 1$
And also:
- $\paren {n + 1}^2 - 1 = n^2 + 2 n$
Combining these results, we obtain:
\(\ds n + 2 \sum_{i \mathop = 1}^n i\) | \(=\) | \(\ds n^2 + 2 n\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 \sum_{i \mathop = 1}^n i\) | \(=\) | \(\ds n \left({n + 1}\right)\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sum_{i \mathop = 1}^n i\) | \(=\) | \(\ds \frac {n \paren {n + 1} } 2\) |
This concludes the proof.
$\blacksquare$