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Question Number 169713 by mathlove last updated on 06/May/22

Commented by Shrinava last updated on 06/May/22

E = 2018

$$\mathrm{E}\:=\:\mathrm{2018} \\ $$

Commented by mathlove last updated on 07/May/22

how is???

$${how}\:{is}??? \\ $$

Commented by mr W last updated on 07/May/22

generally x+n= =(√((x+n)^2 )) =(√(1+(x+n)^2 −1)) =(√(1+(x+n−1)(x+n+1))) =(√(1+(x+n−1)(√(1+(x+n)(x+n+2))))) =(√(1+(x+n−1)(√(1+(x+n)(√(1+(x+n+1)(x+n+3))))))) .... with x=2017, n=1 2018=(√(1+2017(√(1+2018(√(1+2019(√(1+2020(√(1+...))))))))))

$${generally} \\ $$$${x}+{n}= \\ $$$$=\sqrt{\left({x}+{n}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\mathrm{1}+\left({x}+{n}\right)^{\mathrm{2}} −\mathrm{1}} \\ $$$$=\sqrt{\mathrm{1}+\left({x}+{n}−\mathrm{1}\right)\left({x}+{n}+\mathrm{1}\right)} \\ $$$$=\sqrt{\mathrm{1}+\left({x}+{n}−\mathrm{1}\right)\sqrt{\mathrm{1}+\left({x}+{n}\right)\left({x}+{n}+\mathrm{2}\right)}} \\ $$$$=\sqrt{\mathrm{1}+\left({x}+{n}−\mathrm{1}\right)\sqrt{\mathrm{1}+\left({x}+{n}\right)\sqrt{\mathrm{1}+\left({x}+{n}+\mathrm{1}\right)\left({x}+{n}+\mathrm{3}\right)}}} \\ $$$$.... \\ $$$${with}\:{x}=\mathrm{2017},\:{n}=\mathrm{1} \\ $$$$\mathrm{2018}=\sqrt{\mathrm{1}+\mathrm{2017}\sqrt{\mathrm{1}+\mathrm{2018}\sqrt{\mathrm{1}+\mathrm{2019}\sqrt{\mathrm{1}+\mathrm{2020}\sqrt{\mathrm{1}+...}}}}} \\ $$

Commented by mathlove last updated on 07/May/22

thanks mrW

$${thanks}\:{mrW} \\ $$

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