Question Number 152122 by peter frank last updated on 26/Aug/21
$$\sqrt{\mathrm{3}+\sqrt{\mathrm{6}+\sqrt{\mathrm{9}+…+\sqrt{\mathrm{96}+\sqrt{\mathrm{99}}}}}}\:=\:? \\ $$$$ \\ $$
Answered by peter frank last updated on 26/Aug/21
$$\mathrm{y}=\sqrt{\mathrm{3}+\mathrm{y}}\: \\ $$$$\mathrm{y}^{\mathrm{2}} −\mathrm{y}−\mathrm{3}=\mathrm{0}\: \\ $$$$\mathrm{y}=\frac{\mathrm{1}\pm\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$
Commented by MJS_new last updated on 26/Aug/21
$$\mathrm{you}\:\mathrm{solved} \\ $$$${y}=\sqrt{\mathrm{3}+\sqrt{\mathrm{3}+\sqrt{\mathrm{3}+…_{\rightarrow\:\mathrm{to}\:\mathrm{infinity}} }}} \\ $$$$\Rightarrow\:{y}>\mathrm{0} \\ $$$${y}^{\mathrm{2}} =\mathrm{3}+{y} \\ $$$${y}^{\mathrm{2}} −{y}−\mathrm{3}=\mathrm{0}\wedge{y}>\mathrm{0}\:\Rightarrow\:{y}=\frac{\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$$\mathrm{but}\:\mathrm{I}\:\mathrm{think}\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{give}\:\mathrm{a}\:“\mathrm{nice}''\:\mathrm{closed} \\ $$$$\mathrm{expression}\:\mathrm{for} \\ $$$${y}=\sqrt{\mathrm{3}+\sqrt{\mathrm{2}×\mathrm{3}+\sqrt{\mathrm{3}×\mathrm{3}+…+\sqrt{\mathrm{33}×\mathrm{3}}}}} \\ $$
Commented by peter frank last updated on 26/Aug/21
$$\mathrm{thank}\:\mathrm{you} \\ $$