Sum of Sequence of Squares/Proof by Telescoping Series

Proof
Observe that:
 * $3 i \paren {i + 1} = i \paren {i + 1} \paren {i + 2} - i \paren {i + 1} \paren {i - 1}$

That is:
 * $(1): \quad 6 T_i = \paren {i + 1} \paren {\paren {i + 1} + 1} \paren {\paren {i + 1} - 1} - i \paren {i + 1} \paren {i - 1}$

Then:

where $T_n$ is the $n$th Triangular number.

Then: