Difference of Two Powers/Proof 3
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Theorem
Let $\mathbb F$ denote one of the standard number systems, that is $\Z$, $\Q$, $\R$ and $\C$.
Let $n \in \N$ such that $n \ge 2$.
Then for all $a, b \in \mathbb F$:
\(\ds a^n - b^n\) | \(=\) | \(\ds \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \dotsb + a b^{n - 2} + b^{n - 1} }\) |
Proof
From Sum of Geometric Sequence:
\(\ds \sum_{j \mathop = 0}^{n - 1} x^j\) | \(=\) | \(\ds \frac {x^n - 1} {x - 1}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {\dfrac a b}^n - 1\) | \(=\) | \(\ds \paren {\dfrac a b - 1} \sum_{j \mathop = 0}^{n - 1} \paren {\dfrac a b}^j\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {\dfrac {a^n - b^n} {b^n} }\) | \(=\) | \(\ds \paren {\dfrac {a - b} b} \paren {\paren {\dfrac a b}^{n - 1} + \paren {\dfrac a b}^{n - 2} + \dotsb + \paren {\dfrac a b}^1 + 1}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a^n - b^n\) | \(=\) | \(\ds \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + \dotsb + a b^{n - 2} + b^{n - 1} }\) | multiplying both sides by $b^n$ |
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 3$: Natural Numbers: Exercise $\S 3.11 \ (2)$