Menu Close

1-4-16-256-




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

Leave a Reply

Your email address will not be published. Required fields are marked *