|
|
The Purplemath Forums |
Derivation of Finite Geometric Summation Formula This formula is actually quite simple to prove: you just use polynomial long division. The summation formula is:
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 arn–1 + arn–2 + ... + ar3 + ar2 + ar + a A basic property of polynomials is that if you divide xn – 1 by x – 1, you'll get: xn–1 + xn–2 + ... + x3 + x2 + x + 1 That is:
Applying the above to the geometric summation (and reversing both subtractions, so the value of that last fraction isn't changed), we get: arn–1 + arn–2 + ... + ar3 + ar2 + ar + a = a(rn–1 + rn–2 + ... + r3 + r2 + r + 1)
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.
|
|
|
|
Copyright © 2006-2012 Elizabeth Stapel | About | Terms of Use |
|
|
|
|
|
|
![sum[i=1 to n][a-sub-i] = a [ (1 - r^n) / (1 - r) ]](series/sum19.gif){kind=small}
![= a [ (1 - r^n) / (1 - r) ]](series/sum34.gif)
