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Peculiar Sums

A=n=1(1ϕ)n,B=n=0(1ϕ2)n\displaystyle A = \sum_{n=1}^\infty \left ( \dfrac{1}{\phi}\right )^{n} , \qquad B= \displaystyle \sum_{n=0}^\infty \left( \dfrac{1}{\phi^{2}} \right)^{n}

Let ϕ\phi denote the golden ratio, ϕ=1+52\phi = \frac{1+\sqrt5}2 . Then find A+BA+B to 3 decimal places.

Hint: 1ϕ=ϕ1\frac 1\phi = \phi - 1 .

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