Sum of Sequence of Even Squares

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Theorem

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


Proof

\(\ds \sum_{i \mathop = 0}^n \paren {2 i}^2\) \(=\) \(\ds 4 \sum_{i \mathop = 1}^n i^2\) adjustment of indices: $4 i^2 = 0$ when $i = 0$
\(\ds \) \(=\) \(\ds 4 \frac {n \paren {n + 1} \paren {2 n + 1} } 6\) Sum of Sequence of Squares
\(\ds \) \(=\) \(\ds \frac {2 n \paren {n + 1} \paren {2 n + 1} } 3\) simplifying

$\blacksquare$