Difference of Two Even Powers
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Theorem
Let $\GF$ denote one of the standard number systems, that is $\Z$, $\Q$, $\R$ and $\C$.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then for all $a, b \in \GF$:
| \(\ds a^{2 n} - b^{2 n}\) | \(=\) | \(\ds \paren {a - b} \paren {a + b} \sum_{j \mathop = 0}^{n - 1} a^{2 \paren {n - j - 1} } b^{2 j}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {a - b} \paren {a + b} \paren {a^{2 n - 2} + a^{2 n - 4} b^2 + a^{2 n - 6} b^4 + \dotsb + a^2 b^{2 n - 4} + b^{2 n - 2} }\) |
Proof
| \(\ds a^{2 n} - b^{2 n}\) | \(=\) | \(\ds \paren {a^2}^n - \paren {b^2}^n\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {a^2 - b^2} \paren {\sum_{j \mathop = 0}^{n - 1} \paren {a^2} ^{n - j - 1} \paren {b^2}^j}\) | Difference of Two Powers | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {a - b} \paren {a + b} \sum_{j \mathop = 0}^{n - 1} a^{2 \paren {n - j - 1} } b^{2 j}\) | Difference of Two Squares |
whence the result.
$\blacksquare$