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Question Number 152122 by peter frank last updated on 26/Aug/21

$$\sqrt{3+\sqrt{6+\sqrt{9+...+\sqrt{96+\sqrt{99}}}}}=?$$$$$$

Answered by peter frank last updated on 26/Aug/21

$$\mathrm{y}=\sqrt{3+\mathrm{y}}$$$${\mathrm{y}}^2-\mathrm{y}-3=0$$$$\mathrm{y}=\frac{1\pm \sqrt{13}}{2}$$

Commented by MJS_new last updated on 26/Aug/21

$$\mathrm{you}\mathrm{solved}$$$$y=\sqrt{3+\sqrt{3+\sqrt{3+..{.}_{\to \mathrm{to}\mathrm{infinity}}}}}$$$$\Rightarrow y>0$$$$y^2=3+y$$$$y^2-y-3=0\wedge y>0\Rightarrow y=\frac{1+\sqrt{13}}{2}$$$$\mathrm{but}\mathrm{I}\mathrm{think}\mathrm{we}\mathrm{cannot}\mathrm{give}\mathrm{a}‘‘{\mathrm{nice}}^{″}\mathrm{closed}$$$$\mathrm{expression}\mathrm{for}$$$$y=\sqrt{3+\sqrt{2\times 3+\sqrt{3\times 3+...+\sqrt{33\times 3}}}}$$

Commented by peter frank last updated on 26/Aug/21

$$\mathrm{thank}\mathrm{you}$$

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