Closed Form for Triangular Numbers/Proof by Telescoping Sum

Theorem
The closed-form expression for the $n$th triangular number is:
 * $\displaystyle T_n = \sum_{i \mathop = 1}^{n} i = \frac {n \left({n+1}\right)} 2$

Proof
Observe that:

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:

This concludes the proof.