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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: Rearranging the terms of the series into the usual "descending order" for polynomials, we get a series expansion of: Copyright © Elizabeth Stapel 20062011 All Rights Reserved ar^{n}^{–1} + ar^{n}^{–2} + ... + ar^{3} + ar^{2} + ar + a A basic property of polynomials is that if you divide x^{n} – 1 by x – 1, you'll get: x^{n}^{–1} + x^{n}^{–2} + ... + x^{3} + x^{2} + 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: ar^{n}^{–1} + ar^{n}^{–2} + ... + ar^{3} + ar^{2} + ar + a = a(r^{n}^{–1} + r^{n}^{–2} + ... + r^{3} + r^{2} + 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 r^{n} must get closer and closer to zero as n gets larger. Very quickly, r^{n} is as close to nothing as makes no difference, and, "at infinity", is ignored. This is, roughlyspeaking, why the r^{n} is missing in the infinitesum formula.



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