Closed Form for Triangular Numbers/Proof by Telescoping Sum

Theorem

The closed-form expression for the $n$th triangular number is:

$\displaystyle T_n = \sum_{i=1}^{n} i = \frac {n \left({n+1}\right)} 2$

Proof

Observe that:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{i=1}^n \left({ \left({i+1}\right)^2 - i^2 }\right)\)	\(=\)	\(\displaystyle -\sum_{i=1}^n \left({ i^2 - \left({i+1}\right)^2 }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle - \left({1 - \left({n+1}\right)^2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	since this is a telescoping sum
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({n+1}\right)^2 - 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Moreover, we have:

$ \left({i + 1}\right)^2 - i^2 = 2 i + 1$

And also:

$ \left({n + 1}\right)^2 - 1 = n^2 + 2 n$

Combining these results, we obtain:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle n + 2 \sum_{i=1}^n \, i\)	\(=\)	\(\displaystyle n^2 + 2 n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 2 \sum_{i=1}^n \, i\)	\(=\)	\(\displaystyle n \left({n + 1}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \sum_{i=1}^n \, i\)	\(=\)	\(\displaystyle \frac {n \left({n + 1}\right)} 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

This concludes the proof.

$\blacksquare$