Sum of Sequence of Products of Consecutive Integers/Proof 3
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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
| \(\ds \sum_{j \mathop = 1}^n j \paren {j + 1}\) | \(=\) | \(\ds \sum_{j \mathop = 1}^n \paren {j^2 + j}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 1}^n j^2 + \sum_{j \mathop = 1}^n j\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 1}^n j^2 + \frac {n \paren {n + 1} } 2\) | Closed Form for Triangular Numbers | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {n \paren {n + 1} \paren {2 n + 1} } 6 + \frac {n \paren {n + 1} } 2\) | Sum of Sequence of Squares | |||||||||||
| \(\ds \) | \(=\) | \(\ds n \paren {n + 1} \paren {\frac {\paren {2 n + 1} } 6 + \frac 1 2}\) | extracting $n \paren {n + 1}$ as a common factor | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {n \paren {n + 1} \paren {n + 2} } 3\) | simplifying |