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.

Yoink Feb 3, 2019

#1**+1 **

Let r be the common ratio.

The series starts like this: 1, r, r^{2}, r^{3}, ...

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

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 r_{1} by S_{1}.

Denote the sum of the series with common ratio r_{2} by S_{2}.

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

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

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

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

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

MaxWong Feb 4, 2019