Sum of Sequence of Products of Consecutive Integers/Proof 2

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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.

$\blacksquare$