Sum of Sequence of Squares/Proof by Products of Consecutive Integers

Theorem

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

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

$=$

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

$\ds \leadsto \ \ $

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

$=$

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

$\ds \leadsto \ \ $

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

$=$

$\ds n \paren {n + 1} \paren {n + 2} - 3 \frac {n \paren {n + 1} } 2$

$\ds \leadsto \ \ $

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

$=$

$\ds \frac {n \paren {n + 1} \paren {n + 2} } 3 - \frac {n \paren {n + 1} } 2$

$\ds \leadsto \ \ $

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

$=$

$\ds \frac {2 n \paren {n + 1} \paren {n + 2} - 3 n \paren {n + 1} } 6$

$\ds \leadsto \ \ $

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

$=$

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

$\blacksquare$