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

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

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

Thus:
 * $\displaystyle 2 \sum_{i=1}^n i + n = n^2 + 2 n \implies 2 \sum_{i=1}^n i = n \left({n + 1}\right) \implies \sum_{i=1}^n i = \frac {n \left({n + 1}\right)} 2$

as required.