Difference of Two Powers/Proof 1
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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
Let $\ds S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$.
This can also be written:
- $\ds S_n = \sum_{j \mathop = 0}^{n - 1} b^j a^{n - j - 1}$
Consider:
- $\ds a S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j} b^j$
Taking the first term (where $j = 0$) out of the summation, we get:
- $\ds a S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j} b^j = a^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} b^j$
Similarly, consider:
- $\ds b S_n = \sum_{j \mathop = 0}^{n - 1} a^j b^{n - j}$
Taking the first term (where $j = 0$) out of the summation:
- $\ds b S_n = \sum_{j \mathop = 0}^{n - 1} a^j b^{n - j} = b^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} b^j$
This is equal to:
- $\ds b^n + \sum_{j \mathop = 1}^{n - 1} a^j b^{n - j}$
by Permutation of Indices of Summation.
So:
\(\ds \paren {a - b} S_n\) | \(=\) | \(\ds a \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j - b \sum_{j \mathop = 0}^{n - 1} a^j b^{n - j - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 0}^{n - 1} a^{n - j} b^j - \sum_{j \mathop = 0}^{n - 1} a^j b^{n - j}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} b^j - \sum_{j \mathop = 1}^{n - 1} a^{n - j} b^j - b^n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a^n - b^n\) |
$\blacksquare$