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Question Number 153866 by amin96 last updated on 11/Sep/21

$$3+\sqrt{3+\sqrt{6+\sqrt{9+\sqrt{12+\dots +\sqrt{99}}}}}=?$$

Answered by peter frank last updated on 11/Sep/21

$$152122$$

Commented by amin96 last updated on 11/Sep/21

$$$$how did you do the solution sir?

Commented by mr W last updated on 11/Sep/21

$$i{don}^{\prime }tthinkitcanbesolvedinclosed$$$$form.$$$$nomatterhowyougottheresult152122,$$$$theresultiswrong.$$$$$$$$S=3+\sqrt{3+\sqrt{6+\sqrt{9+\sqrt{....+\sqrt{99}}}}}$$$$<3+\sqrt{99+\sqrt{99+\sqrt{99+\sqrt{....+\sqrt{99}}}}}$$$$<3+\sqrt{99+\sqrt{99+\sqrt{99+\sqrt{....+\sqrt{99+\sqrt{...}}}}}}$$$$=3+L$$$$withL=\sqrt{99+\sqrt{99+\sqrt{99+\sqrt{....+\sqrt{99+\sqrt{...}}}}}}$$$$L=\sqrt{99+L}$$$$L^2-L-99=0$$$$L=\frac{1+\sqrt{397}}{2}\approx 10.5$$$$\Rightarrow 3+\sqrt{3+\sqrt{6+\sqrt{9+\sqrt{....+\sqrt{99}}}}}<3+L=13.5$$$$soitcanneverbe152122!$$

Commented by MJS_new last updated on 11/Sep/21

$$\mathrm{can}\mathrm{we}\mathrm{find}\mathrm{a}\mathrm{closed}\mathrm{form}\mathrm{if}\mathrm{it}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{stop}?$$$$\sqrt{1\times 3+\sqrt{2\times 3+\sqrt{3\times 3+...\sqrt{n\times 3}}}}\mathrm{with}n\to \mathrm{\infty }$$$$\mathrm{or}\mathrm{at}\mathrm{least}\mathrm{for}$$$$\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+...\sqrt{n}}}}}\mathrm{with}n\to \mathrm{\infty }$$

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