Sum of Sequence of Squares/Proof by Summation of Summations

Theorem

$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$

We can observe from the above diagram that:

$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \sum_{i \mathop = 1}^n \paren {\sum_{j \mathop = i}^n j}$

Therefore we have:

$\ds \sum_{i \mathop = 1}^n i^2$

$=$

$\ds \sum_{i \mathop = 1}^n \paren {\sum_{j \mathop = i}^n j}$

$\ds $

$=$

$\ds \sum_{i \mathop = 1}^n \paren {\sum_{j \mathop = 1}^n j - \sum_{j \mathop = 1}^{i - 1} j}$

$\ds $

$=$

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

$\ds \leadsto \ \ $

$\ds 2 \sum_{i \mathop = 1}^n i^2$

$=$

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

$\ds \leadsto \ \ $

$\ds 3 \sum_{i \mathop = 1}^n i^2$

$=$

$\ds n^2 \paren {n + 1} + \sum_{i \mathop = 1}^n i$

$\ds \leadsto \ \ $

$\ds 3 \sum_{i \mathop = 1}^n i^2$

$=$

$\ds n^2 \paren {n + 1} + \frac {n \paren {n + 1} } 2$

$\ds \leadsto \ \ $

$\ds 6 \sum_{i \mathop = 1}^n i^2$

$=$

$\ds 2 n^2 \paren {n + 1} + n \paren {n + 1}$

$\ds $

$=$

$\ds n \paren {n + 1} \paren {2 n + 1}$

$\ds \leadsto \ \ $

$\ds \sum_{i \mathop = 1}^n i^2$

$=$

$\ds \frac {n \paren {n + 1} \paren {2 n + 1} } 6$

$\blacksquare$