Sum of Sequence of Odd Squares/Formulation 1

Theorem

$\ds \forall n \in \N: \sum_{i \mathop = 0}^n \paren {2 i + 1}^2 = \frac {\paren {n + 1} \paren {2 n + 1} \paren {2 n + 3} } 3$

$\ds \sum_{i \mathop = 0}^n \paren {2 i + 1}^2$

$=$

$\ds \sum_{i \mathop = 0}^n \paren {2 i}^2 + \sum_{i \mathop = 0}^n 4 i + \sum_{i \mathop = 0}^n 1$

$\ds $

$=$

$\ds \frac {2 n \paren {n + 1} \paren {2 n + 1} } 3 + 4 \sum_{i \mathop = 0}^n i + \sum_{i \mathop = 0}^n 1$

$\ds $

$=$

$\ds \frac {2 n \paren {n + 1} \paren {2 n + 1} } 3 + 4 \sum_{i \mathop = 1}^n i + \sum_{i \mathop = 0}^n 1$

adjustment of indices: $4 i = 0$ when $i = 0$

$\ds $

$=$

$\ds \frac {2 n \paren {n + 1} \paren {2 n + 1} } 3 + 4 \frac {n \paren {n + 1} } 2 + \sum_{i \mathop = 0}^n 1$

$\ds $

$=$

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

further simplification

$\ds $

$=$

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

factorising

$\ds $

$=$

$\ds \frac {\paren {n + 1} \paren {4 n^2 + 8 n + 3} } 3$

multiplying out

$\ds $

$=$

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

factorising

$\blacksquare$