Difference of Two Powers/General Commutative Ring
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Theorem
Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$.
Let $a, b \in R$.
Let $n \in \N$ such that $n \ge 2$.
Then:
\(\ds a^n - b^n\) | \(=\) | \(\ds \paren {a - b} \circ \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} \circ b^j\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a - b} \circ \paren {a^{n - 1} + a^{n - 2} \circ b + a^{n - 3} \circ b^2 + \dotsb + a \circ b^{n - 2} + b^{n - 1} }\) |
Proof
Let $\ds S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} \circ b^j$.
This can also be written:
- $\ds S_n = \sum_{j \mathop = 0}^{n - 1} b^j \circ a^{n - j - 1}$
Consider:
- $\ds a \circ S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j} \circ b^j$
Taking the first term (where $j = 0$) out of the summation, we get:
- $\ds a \circ S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j} \circ b^j = a^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} \circ b^j$
Similarly, consider:
- $\ds b \circ S_n = \sum_{j \mathop = 0}^{n - 1} a^j \circ b^{n - j}$
Taking the first term (where $j = 0$) out of the summation:
- $\ds b \circ S_n = \sum_{j \mathop = 0}^{n - 1} a^j \circ b^{n - j} = b^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} \circ b^j$
This is equal to:
- $\ds b^n + \sum_{j \mathop = 1}^{n - 1} a^j \circ b^{n - j}$
by Permutation of Indices of Summation.
So:
\(\ds \paren {a - b} \circ S_n\) | \(=\) | \(\ds a \circ \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} \circ b^j - b \circ \sum_{j \mathop = 0}^{n - 1} a^j \circ b^{n - j - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 0}^{n - 1} a^{n - j} \circ b^j - \sum_{j \mathop = 0}^{n - 1} a^j \circ b^{n - j}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} \circ b^j - \sum_{j \mathop = 1}^{n - 1} a^{n - j} \circ b^j - b^n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a^n - b^n\) |
$\blacksquare$