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Question Number 24355 by ahmetbak1r last updated on 16/Nov/17

(√(1+(√(4+(√(16+(√(256.....))))))))=?

$$\sqrt{\mathrm{1}+\sqrt{\mathrm{4}+\sqrt{\mathrm{16}+\sqrt{\mathrm{256}.....}}}}=? \\ $$

Answered by ajfour last updated on 16/Nov/17

S=(√(2^0 +(√(2^2 +(√(2^4 +(√(2^8 +(√(..)))))))))) S^( 2) =1+2(√(1+(√(1+(√(1+(√(..)))))))) S^( 2) = 1+2E , where E = (√(1+(√(1+(√(1+(√(..)))))))) ⇒ E^( 2) =1+E or E^( 2) −E=1 with E >1 (E−(1/2))^2 =(5/4) E =(1/2)+((√5)/2) As S^( 2) =2E+1 S^( 2) =(√5)+2 ⇒ S = (√(2+(√5))) .

$${S}=\sqrt{\mathrm{2}^{\mathrm{0}} +\sqrt{\mathrm{2}^{\mathrm{2}} +\sqrt{\mathrm{2}^{\mathrm{4}} +\sqrt{\mathrm{2}^{\mathrm{8}} +\sqrt{..}}}}} \\ $$$${S}^{\:\mathrm{2}} =\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{..}}}} \\ $$$${S}^{\:\mathrm{2}} =\:\mathrm{1}+\mathrm{2}{E}\:\:\:,\:{where} \\ $$$${E}\:=\:\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{..}}}} \\ $$$$\Rightarrow\:\:\:{E}^{\:\mathrm{2}} \:=\mathrm{1}+{E} \\ $$$${or}\:\:\:{E}^{\:\mathrm{2}} −{E}=\mathrm{1}\:\:\:\:{with}\:{E}\:>\mathrm{1} \\ $$$$\left({E}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${E}\:=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${As}\:\:\:\:\:\:\:{S}^{\:\mathrm{2}} \:=\mathrm{2}{E}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{S}^{\:\mathrm{2}} \:=\sqrt{\mathrm{5}}+\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:{S}\:=\:\sqrt{\mathrm{2}+\sqrt{\mathrm{5}}}\:. \\ $$

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