Derivation of the Geometric Summation Formula
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The formula for the n-th partial sum, Sn, of a geometric series with common ratio r is given by:
![sum[i=1 to n][a-sub-i] = a [ (1 − r^n) / (1 − r) ]](series/sum19.gif){kind=small}
This formula is actually quite simple to confirm: you just use polynomial long division.
The sum of the first n terms of the geometric sequence, in expanded form, is as follows:
a + ar + ar2 + ar3 + ... + arn−2 + arn−1
Polynomials are customarily written with their terms in "descending order". Reversing the order of the summation above to put its terms in descending order, we get a series expansion of:
arn−1 + arn−2 + ... + ar3 + ar2 + ar + a
We can take the common factor of "a" out front:
a(rn−1 + rn−2 + ... + r3 + r2 + r + 1)
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
This means that:
If we reverse both subtractions in the fraction above, we will obtain the following equivalent equation:
Applying the above to the geometric summation (by using "r" instead of "x"), we get:
![= a [ (1 − r^n) / (1 − r) ]](series/sum34.gif)
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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