# Closed Form for Triangular Numbers/Proof by Telescoping Sum

## 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.

$\blacksquare$