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Question Number 149423 by mathdanisur last updated on 05/Aug/21

((2207 - (1/(2207 - (1/(2207 - ...))))))^(1/8)   = ?

$$\sqrt[{\mathrm{8}}]{\mathrm{2207}\:-\:\frac{\mathrm{1}}{\mathrm{2207}\:-\:\frac{\mathrm{1}}{\mathrm{2207}\:-\:...}}}\:\:=\:? \\ $$

Answered by dumitrel last updated on 05/Aug/21

x=2207−(1/(2207−(1/(2207−....))))⇒x=2207−(1/x)⇒x=((2207+21∙47(√5))/2)  (((1+(√5))/2))^(16) =...=((2^(15) (2207+21∙47(√5)))/2^(16) )=x⇒(((((1+(√5))/2))^(16) ))^(1/8) =(((1+(√5))/2))^2 =((3+(√5))/2)

$${x}=\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{2207}−....}}\Rightarrow{x}=\mathrm{2207}−\frac{\mathrm{1}}{{x}}\Rightarrow{x}=\frac{\mathrm{2207}+\mathrm{21}\centerdot\mathrm{47}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{16}} =...=\frac{\mathrm{2}^{\mathrm{15}} \left(\mathrm{2207}+\mathrm{21}\centerdot\mathrm{47}\sqrt{\mathrm{5}}\right)}{\mathrm{2}^{\mathrm{16}} }={x}\Rightarrow\sqrt[{\mathrm{8}}]{\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{16}} }=\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

Commented by mathdanisur last updated on 05/Aug/21

Thank You Ser, Cool

$${Thank}\:{You}\:{Ser},\:{Cool} \\ $$

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