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 |
$\blacksquare$