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In a certain infinite geometric series, the first term is 1, and each term is equal to the sum of the two terms after it. (For example, the first term is equal to the sum of the second term and third term). If S is the sum of the series and R is the common ratio, find S-R.

 Feb 3, 2019
edited by Yoink  Feb 3, 2019
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Let r be the common ratio.

The series starts like this: 1, r, r2, r3, ...

\(r^n = r^{n+1} + r^{n+2}\\ r^2 + r - 1 = 0\\ r = \dfrac{-1\pm\sqrt{5}}{2}\)

Now we found the possible common ratios. 

Denote \(r_1 = \dfrac{-1-\sqrt5}{2},\;r_2 = \dfrac{\sqrt 5 -1}{2}\)

Denote the sum of the series with common ratio r1 by S1.

Denote the sum of the series with common ratio r2 by S2.

\(S_1 = \dfrac{1}{1-\dfrac{-1-\sqrt 5}{2}} = \dfrac{3-\sqrt 5}{2}\)

\(S_2 = \dfrac{1}{1-\dfrac{\sqrt5-1}{2}} = \dfrac{3+\sqrt5}{2}\)

\(S_1 - r_1 = \dfrac{3-\sqrt5}{2} - \dfrac{-1-\sqrt5}{2} = 2\)

\(S_2 - r_2 = \dfrac{3+\sqrt5}{2} - \dfrac{\sqrt5 - 1}{2} = 2\)

 

For both cases, we get S-R = 2.

 Feb 4, 2019

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