# Sum of Sequence of Products of Consecutive Integers/Proof 2

## Theorem

 $\ds \sum_{j \mathop = 1}^n j \paren {j + 1}$ $=$ $\ds 1 \times 2 + 2 \times 3 + \dotsb + n \paren {n + 1}$ $\ds$ $=$ $\ds \frac {n \paren {n + 1} \paren {n + 2} } 3$

## Proof

Observe that:

 $\ds 3 i \paren {i + 1}$ $=$ $\ds i \paren {i + 1} \paren {i + 2} - i \paren {i + 1} \paren {i - 1}$ $\ds$ $=$ $\ds \paren {i + 1} \paren {\paren {i + 1} + 1} \paren {\paren {i + 1} - 1} - i \paren {i + 1} \paren {i - 1}$

Thus:

$3 i \paren {i + 1} = b \paren {i + 1} - b \paren i$

where:

$b \paren i = i \paren {i + 1} \paren {i - 1}$

This can therefore be used as the basis of a telescoping series, as follows:

 $\ds \sum_{i \mathop = 1}^n 3 i \paren {i + 1}$ $=$ $\ds \sum_{i \mathop = 1}^n \paren {i \paren {i + 1} \paren {i + 2} - i \paren {i + 1} \paren {i - 1} }$ $\ds$ $=$ $\ds n \paren {n + 1} \paren {n + 2} - 1 \paren {1 + 1} \paren {1 - 1}$ Telescoping Series $\ds$ $=$ $\ds n \paren {n + 1} \paren {n + 2}$

Hence the result.

$\blacksquare$