Sum of Sequence of Squares/Proof by Sum of Differences of Cubes
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Theorem
- $\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$
Proof
| \(\ds \sum_{i \mathop = 1}^n \paren {\paren {i + 1}^3 - i^3}\) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \paren {i^3 + 3 i^2 + 3 i + 1 - i^3}\) | Binomial Theorem | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \paren {3 i^2 + 3 i + 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 3 \sum_{i \mathop = 1}^n i^2 + 3 \sum_{i \mathop = 1}^n i + \sum_{i \mathop = 1}^n 1\) | Summation is Linear | |||||||||||
| \(\ds \) | \(=\) | \(\ds 3\sum_{i \mathop = 1}^n i^2 + 3 \frac {n \paren {n + 1} } 2 + n\) | Closed Form for Triangular Numbers |
On the other hand:
| \(\ds \sum_{i \mathop = 1}^n \paren {\paren {i + 1}^3 - i^3}\) | \(=\) | \(\ds \paren {n + 1}^3 - n^3 + n^3 - \paren {n - 1}^3 + \paren {n - 1}^3 - \cdots + 2^3 - 1^3\) | Definition of Summation | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {n + 1}^3 - 1^3\) | Telescoping Series: Example 2 | |||||||||||
| \(\ds \) | \(=\) | \(\ds n^3 + 3n^2 + 3n + 1 - 1\) | Binomial Theorem | |||||||||||
| \(\ds \) | \(=\) | \(\ds n^3 + 3n^2 + 3n\) |
Therefore:
| \(\ds 3 \sum_{i \mathop = 1}^n i^2 + 3 \frac {n \paren {n + 1} } 2 + n\) | \(=\) | \(\ds n^3 + 3 n^2 + 3 n\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 3 \sum_{i \mathop = 1}^n i^2\) | \(=\) | \(\ds n^3 + 3 n^2 + 3 n - 3 \frac {n \paren {n + 1} } 2 - n\) |
Therefore:
| \(\ds \sum_{i \mathop = 1}^n i^2\) | \(=\) | \(\ds \frac 1 3 \paren {n^3 + 3 n^2 + 3 n - 3 \frac {n \paren {n + 1} }2 - n}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 3 \paren {n^3 + 3 n^2 + 3 n - \frac {3 n^2} 2 - \frac {3 n} 2 - n}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 3 \paren {n^3 + \frac {3 n^2} 2 + \frac n 2}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 6 n \paren {2 n^2 + 3 n + 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 6 n \paren {n + 1} \paren {2 n + 1}\) |
$\blacksquare$