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:

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


Proof

Observe that:

\(\displaystyle \sum_{i \mathop = 1}^n \paren {\paren {i + 1}^2 - i^2}\) \(=\) \(\displaystyle -\sum_{i \mathop = 1}^n \paren {i^2 - \paren {i + 1} ^2}\)
\(\displaystyle \) \(=\) \(\displaystyle -\paren {1 - \paren {n + 1}^2}\) Telescoping Series
\(\displaystyle \) \(=\) \(\displaystyle \paren {n + 1}^2 - 1\)

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:

\(\displaystyle n + 2 \sum_{i \mathop = 1}^n i\) \(=\) \(\displaystyle n^2 + 2 n\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 2 \sum_{i \mathop = 1}^n i\) \(=\) \(\displaystyle n \left({n + 1}\right)\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{i \mathop = 1}^n i\) \(=\) \(\displaystyle \frac {n \paren {n + 1} } 2\)

This concludes the proof.

$\blacksquare$