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8-8-2-8-4-8-8-8-16-




Question Number 204664 by cortano12 last updated on 25/Feb/24
  8+(√(8^2 +(√(8^4 +(√(8^8 +(√(8^(16) +(√(...)))))))))) = ?
$$\:\:\mathrm{8}+\sqrt{\mathrm{8}^{\mathrm{2}} +\sqrt{\mathrm{8}^{\mathrm{4}} +\sqrt{\mathrm{8}^{\mathrm{8}} +\sqrt{\mathrm{8}^{\mathrm{16}} +\sqrt{…}}}}}\:=\:?\: \\ $$
Answered by Frix last updated on 25/Feb/24
=8+8(√(1+(√(1+(√(1+(√(1+...))))))))  =8+8ϕ=12+4(√5)
$$=\mathrm{8}+\mathrm{8}\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+…}}}} \\ $$$$=\mathrm{8}+\mathrm{8}\varphi=\mathrm{12}+\mathrm{4}\sqrt{\mathrm{5}} \\ $$
Commented by cortano12 last updated on 25/Feb/24
ϕ = golden ratio
$$\varphi\:=\:\mathrm{golden}\:\mathrm{ratio}\: \\ $$
Answered by SEKRET last updated on 25/Feb/24
  8+(√(8∙8+8(√(8^2 +(√(8^4 +(√(8^8 +....   )))))))) = X   8+(√8) ∙(√(8+(√(8^2 +(√(8^4 +...^(  ) )))) )) = X    8+(√(8 )) ∙(√X) = X       X= 12+4(√5)
$$\:\:\mathrm{8}+\sqrt{\mathrm{8}\centerdot\mathrm{8}+\mathrm{8}\sqrt{\mathrm{8}^{\mathrm{2}} +\sqrt{\mathrm{8}^{\mathrm{4}} +\sqrt{\mathrm{8}^{\mathrm{8}} +….\:\:\:}}}}\:=\:\boldsymbol{\mathrm{X}} \\ $$$$\:\mathrm{8}+\sqrt{\mathrm{8}}\:\centerdot\sqrt{\mathrm{8}+\sqrt{\mathrm{8}^{\mathrm{2}} +\sqrt{\mathrm{8}^{\mathrm{4}} +…^{\:\:} }}\:}\:=\:\boldsymbol{\mathrm{X}} \\ $$$$\:\:\mathrm{8}+\sqrt{\mathrm{8}\:}\:\centerdot\sqrt{\boldsymbol{\mathrm{X}}}\:=\:\boldsymbol{\mathrm{X}} \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{X}}=\:\mathrm{12}+\mathrm{4}\sqrt{\mathrm{5}} \\ $$$$\:\:\:\: \\ $$

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