Question Number 218673 by Spillover last updated on 14/Apr/25
Commented by Nicholas666 last updated on 14/Apr/25
$$\varphi \\ $$
Answered by Nicholas666 last updated on 14/Apr/25
$$\:\frac{\varphi}{\varphi^{\mathrm{2}} −\mathrm{1}}+\frac{\varphi^{\mathrm{2}} }{\varphi^{\mathrm{4}} −\mathrm{1}\:}+\frac{\varphi^{\mathrm{4}} }{\varphi^{\mathrm{8}} −\mathrm{1}}+….. \\ $$$${solution}; \\ $$$$\:\frac{\varphi^{\mathrm{2}{n}−\mathrm{1}} }{\varphi^{\mathrm{2}{n}} −\mathrm{1}}=\:\frac{\mathrm{1}}{\varphi^{\mathrm{2}{n}} −\mathrm{1}}\:−\frac{\mathrm{1}}{\varphi^{\mathrm{2}{n}} −\mathrm{1}} \\ $$$${S}=\left(\frac{\mathrm{1}}{\varphi−\mathrm{1}}\:\:−\:\frac{\mathrm{1}}{\varphi−\mathrm{1}}\right)+\left(\frac{\mathrm{1}}{\varphi^{\mathrm{2}} −\mathrm{1}}\:−\frac{\mathrm{1}}{\varphi^{\mathrm{4}} −\mathrm{1}}\right)+\left(\frac{\mathrm{1}}{\varphi^{\mathrm{4}} −\mathrm{1}}\:−\frac{\mathrm{1}}{\varphi^{\mathrm{8}} −\mathrm{1}}\right)+……\:\: \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\varphi−\mathrm{1}}\:−\:\frac{\mathrm{1}}{\varphi^{\mathrm{2}{n}} −\mathrm{1}} \\ $$$${S}_{{n}\:} =\varphi−\:\frac{\mathrm{1}}{\varphi^{\mathrm{2}{n}} −\mathrm{1}}\:=\varphi\: \\ $$
Answered by Charleston last updated on 14/Apr/25
$$ \\ $$
Answered by Spillover last updated on 14/Apr/25
