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.
$\blacksquare$