Sum of Sequence of Products of Consecutive Integers/Proof 2

Theorem

 * $\displaystyle \sum_{j \mathop = 1}^n j \left({j+1}\right) = \frac {n \left({n+1}\right) \left({n+2}\right)} 3$

Proof
Observe that:

Thus:
 * $3 i \left({i + 1}\right) = b \left({i + 1}\right) - b \left({i}\right)$

where:
 * $b \left({i}\right) = i \left({i + 1}\right) \left({i - 1}\right)$

This can therefore be used as the basis of a Telescoping Series, as follows:

Hence the result.