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Derivation of Finite Geometric Summation Formula


This formula is actually quite simple to prove: you just use polynomial long division.

The summation formula is:

    sum[i=1 to n][a-sub-i] = a [ (1 - r^n) / (1 - r) ]

Rearranging the terms of the series into the usual "descending order" for polynomials, we get a series expansion of: Copyright Elizabeth Stapel 2006-2011 All Rights Reserved

    arn1 + arn2 + ... + ar3 + ar2 + ar + a

A basic property of polynomials is that if you divide xn 1 by x 1, you'll get:

    xn1 + xn2 + ... + x3 + x2 + x + 1

That is:

    x^(n-1) + ... + x^2 + x + 1 = (x^n - 1)/(x - 1)

Applying the above to the geometric summation (and reversing both subtractions, so the value of that last fraction isn't changed), we get:

    arn1 + arn2 + ... + ar3 + ar2 + ar + a

      = a(rn1 + rn2 + ... + r3 + r2 + r + 1)

      = a [ (1 - r^n) / (1 - r) ]

The above derivation can be extended to give the formula for infinite series, but requires tools from calculus. For now, just note that, for r | < 1, a basic property of exponential functions is that rn must get closer and closer to zero as n gets larger. Very quickly, rn is as close to nothing as makes no difference, and, "at infinity", is ignored. This is, roughly-speaking, why the rn is missing in the infinite-sum formula.

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Cite this article as:

Stapel, Elizabeth. "Derivation of Finite Geometric Summation Formula." Purplemath. Available from
    http://www.purplemath.com/modules/series7.htm. Accessed
 

 



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