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Question Number 798 by 123456 last updated on 15/Mar/15

(√(1+2(√(1+3(√(1+4(√(1+∙∙∙))))))))

$$\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+\centerdot\centerdot\centerdot}}}} \\ $$

Answered by prakash jain last updated on 15/Mar/15

Ramanujan′s Formula  x+1=(√(1+x(√(1+(x+1)(√(1+(x+2)(√(1+...))))))))  For x=2  3=(√(1+2(√(1+3(√(1+4(√(1+..))))))))

$$\mathrm{Ramanujan}'{s}\:\mathrm{Formula} \\ $$$${x}+\mathrm{1}=\sqrt{\mathrm{1}+{x}\sqrt{\mathrm{1}+\left({x}+\mathrm{1}\right)\sqrt{\mathrm{1}+\left({x}+\mathrm{2}\right)\sqrt{\mathrm{1}+...}}}} \\ $$$$\mathrm{For}\:{x}=\mathrm{2} \\ $$$$\mathrm{3}=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+..}}}} \\ $$

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