Question-145918

Question Number 145918 by mathdanisur last updated on 09/Jul/21

Answered by Olaf_Thorendsen last updated on 09/Jul/21

l = 3+(√(3^2 +(√(3^4 +(√(3^8 +(√(...)))))))) (l−3)^2 = 3^2 +(√(3^4 +(√(3^8 +(√(...)))))) (l−3)^2 = 3^2 +(√(3^4 +3^2 (√(3^4 +(√(...)))))) (l−3)^2 = 3^2 +3(√(3^2 +(√(3^4 +(√(...)))))) (((l−3)^2 )/3) = 3+(√(3^2 +(√(3^4 +(√(...)))))) = l l^2 −9l+9 = 0 l = ((9±(√(81−4×1×9)))/2) l = ((9±3(√5))/2) ((9−3(√5))/2) ≈ 1,146 : impossible because l>3 Finally l = ((9+3(√5))/2) ≈ 7,854

$${l}\:=\:\mathrm{3}+\sqrt{\mathrm{3}^{\mathrm{2}} +\sqrt{\mathrm{3}^{\mathrm{4}} +\sqrt{\mathrm{3}^{\mathrm{8}} +\sqrt{…}}}} \\ $$$$\left({l}−\mathrm{3}\right)^{\mathrm{2}} \:=\:\mathrm{3}^{\mathrm{2}} +\sqrt{\mathrm{3}^{\mathrm{4}} +\sqrt{\mathrm{3}^{\mathrm{8}} +\sqrt{…}}} \\ $$$$\left({l}−\mathrm{3}\right)^{\mathrm{2}} \:=\:\mathrm{3}^{\mathrm{2}} +\sqrt{\mathrm{3}^{\mathrm{4}} +\mathrm{3}^{\mathrm{2}} \sqrt{\mathrm{3}^{\mathrm{4}} +\sqrt{…}}} \\ $$$$\left({l}−\mathrm{3}\right)^{\mathrm{2}} \:=\:\mathrm{3}^{\mathrm{2}} +\mathrm{3}\sqrt{\mathrm{3}^{\mathrm{2}} +\sqrt{\mathrm{3}^{\mathrm{4}} +\sqrt{…}}} \\ $$$$\frac{\left({l}−\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{3}}\:=\:\mathrm{3}+\sqrt{\mathrm{3}^{\mathrm{2}} +\sqrt{\mathrm{3}^{\mathrm{4}} +\sqrt{…}}}\:=\:{l} \\ $$$${l}^{\mathrm{2}} −\mathrm{9}{l}+\mathrm{9}\:=\:\mathrm{0} \\ $$$${l}\:=\:\frac{\mathrm{9}\pm\sqrt{\mathrm{81}−\mathrm{4}×\mathrm{1}×\mathrm{9}}}{\mathrm{2}} \\ $$$${l}\:=\:\frac{\mathrm{9}\pm\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\frac{\mathrm{9}−\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}\:\approx\:\mathrm{1},\mathrm{146}\::\:\mathrm{impossible}\:\mathrm{because}\:{l}>\mathrm{3} \\ $$$$\mathrm{Finally}\:{l}\:=\:\frac{\mathrm{9}+\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}\:\approx\:\mathrm{7},\mathrm{854}\: \\ $$

Commented by Mrsof last updated on 09/Jul/21