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Question Number 100986 by Dwaipayan Shikari last updated on 29/Jun/20

Commented by bachamohamed last updated on 29/Jun/20

$$\mathit{l}\mathit{o}\mathit{o}\mathit{k}\mathit{t}\mathit{h}\mathit{i}\mathit{s}$$$$2=\sqrt{1+\sqrt{9}}=\sqrt{1+\sqrt{1+8}}=\sqrt{1+\sqrt{1+\sqrt{64}}}=\sqrt{1+\sqrt{1+\sqrt{1+63}}}$$$$=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{{63}^2}}}}=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+({63}^2-1)}}}}$$$$=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{{\left(({63}^2-1)\right)}^2}}}}}=\sqrt{1+\sqrt{1+\sqrt{1.....+\sqrt{1........\mathrm{\infty }}}}}$$$$3=\sqrt{9}=\sqrt{1+\sqrt{64}}=\sqrt{1+\sqrt{1+63}}=\sqrt{1+\sqrt{1+\sqrt{{63}^2}}}=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1......\mathrm{\infty }}}}}$$$$4=\sqrt{16}=\sqrt{1+15}=\sqrt{1+\sqrt{225}}=\sqrt{1+\sqrt{1+224}}=\sqrt{1+\sqrt{1+\sqrt{{224}^2}}}=\sqrt{1+\sqrt{1+\sqrt{1...\sqrt{....\sqrt{1...\mathrm{\infty }}}}}}$$$$\mathrm{donc}\mathrm{pour}\mathrm{tout}\mathrm{nombre}\mathrm{posituve}\mathit{n}$$$$\mathit{n}=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1.....\mathrm{\infty }}}}}$$$$\mathit{w}\mathit{h}\mathit{y}?$$

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