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Question Number 218673 by Spillover last updated on 14/Apr/25

Commented by Nicholas666 last updated on 14/Apr/25

ϕ

$$\varphi \\ $$

Answered by Nicholas666 last updated on 14/Apr/25

(ϕ/(ϕ^2 −1))+(ϕ^2 /(ϕ^4 −1 ))+(ϕ^4 /(ϕ^8 −1))+..... solution; (ϕ^(2n−1) /(ϕ^(2n) −1))= (1/(ϕ^(2n) −1)) −(1/(ϕ^(2n) −1)) S=((1/(ϕ−1)) − (1/(ϕ−1)))+((1/(ϕ^2 −1)) −(1/(ϕ^4 −1)))+((1/(ϕ^4 −1)) −(1/(ϕ^8 −1)))+...... S_n =(1/(ϕ−1)) − (1/(ϕ^(2n) −1)) S_(n ) =ϕ− (1/(ϕ^(2n) −1)) =ϕ

$$\:\frac{\varphi}{\varphi^{\mathrm{2}} −\mathrm{1}}+\frac{\varphi^{\mathrm{2}} }{\varphi^{\mathrm{4}} −\mathrm{1}\:}+\frac{\varphi^{\mathrm{4}} }{\varphi^{\mathrm{8}} −\mathrm{1}}+..... \\ $$$${solution}; \\ $$$$\:\frac{\varphi^{\mathrm{2}{n}−\mathrm{1}} }{\varphi^{\mathrm{2}{n}} −\mathrm{1}}=\:\frac{\mathrm{1}}{\varphi^{\mathrm{2}{n}} −\mathrm{1}}\:−\frac{\mathrm{1}}{\varphi^{\mathrm{2}{n}} −\mathrm{1}} \\ $$$${S}=\left(\frac{\mathrm{1}}{\varphi−\mathrm{1}}\:\:−\:\frac{\mathrm{1}}{\varphi−\mathrm{1}}\right)+\left(\frac{\mathrm{1}}{\varphi^{\mathrm{2}} −\mathrm{1}}\:−\frac{\mathrm{1}}{\varphi^{\mathrm{4}} −\mathrm{1}}\right)+\left(\frac{\mathrm{1}}{\varphi^{\mathrm{4}} −\mathrm{1}}\:−\frac{\mathrm{1}}{\varphi^{\mathrm{8}} −\mathrm{1}}\right)+......\:\: \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\varphi−\mathrm{1}}\:−\:\frac{\mathrm{1}}{\varphi^{\mathrm{2}{n}} −\mathrm{1}} \\ $$$${S}_{{n}\:} =\varphi−\:\frac{\mathrm{1}}{\varphi^{\mathrm{2}{n}} −\mathrm{1}}\:=\varphi\: \\ $$

Answered by Charleston last updated on 14/Apr/25

$$ \\ $$

Answered by Spillover last updated on 14/Apr/25

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