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

$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$
 $\ds \sum_{i \mathop = 1}^n 3 i \paren {i + 1}$ $=$ $\ds n \paren {n + 1} \paren {n + 2}$ Sum of Sequence of Products of Consecutive Integers $\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 \ \$ $\ds \sum_{i \mathop = 1}^n 3 i^2$ $=$ $\ds n \paren {n + 1} \paren {n + 2} - 3 \frac {n \paren {n + 1} } 2$ Closed Form for Triangular Numbers $\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 \ \$ $\ds \sum_{i \mathop = 1}^n i^2$ $=$ $\ds \frac {n \paren {n + 1} \paren {2 n + 1} } 6$
$\blacksquare$