Math 2.0

Combinatorics Theorem

Sum of a finite geometric progression

Builds on Mathematical induction

Statement

Let xx be a real number with x≠1x \neq 1, and let nn be a natural number. Then ∑k=0n−1xk=1+x+x2+⋯+xn−1=xn−1x−1.\sum_{k=0}^{n-1} x^k = 1 + x + x^2 + \dots + x^{n-1} = \frac{x^n - 1}{x - 1}.

Remarks

The same formula holds for xx in any field, for example for complex numbers, when x≠1x \neq 1. For x=1x = 1, the sum is nn. A short proof: (x−1)(1+x+⋯+xn−1)=xn−1(x - 1)(1 + x + \dots + x^{n-1}) = x^n - 1, because the other terms cancel.

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Where this leads

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Further reading

  1. Web https://en.wikipedia.org/wiki/Geometric_progression, Geometric series
About this page
Tag03X8V7D64M
URIm2://03X8V7D64M
Aliasesfinite-geometric-series
Kindstatement, theorem
WritingW0 stub
MSC05A19
StatusThis copy of the site has no build data. Run m2 status 03X8V7D64M.
UsesMathematical induction

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