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Question Number 60577 by naka3546 last updated on 22/May/19

2(√(1 + 3(√(1 + 5(√(1 + 7(√(1 + 11(√(1 + 13(√(1 + 17(√(...)))))))))))))) = x x = ?

$$\mathrm{2}\sqrt{\mathrm{1}\:+\:\mathrm{3}\sqrt{\mathrm{1}\:+\:\mathrm{5}\sqrt{\mathrm{1}\:+\:\mathrm{7}\sqrt{\mathrm{1}\:+\:\mathrm{11}\sqrt{\mathrm{1}\:+\:\mathrm{13}\sqrt{\mathrm{1}\:+\:\mathrm{17}\sqrt{...}}}}}}}\:\:=\:\:{x} \\ $$$${x}\:\:=\:\:? \\ $$

Commented by ajfour last updated on 22/May/19

Is there such a question in a book?

$$\mathrm{Is}\:\mathrm{there}\:\mathrm{such}\:\mathrm{a}\:\mathrm{question}\:\mathrm{in}\:\mathrm{a}\:\mathrm{book}? \\ $$

Answered by Prithwish sen last updated on 22/May/19

(n+1)(n+3)=(n+1)(√((n+3)^2 )) =(n+1)(√(1+(n+2)(n+4))) =(n+1)(√(1+(n+2)(√((n+4)^2 )))) =(n+1)(√(1+(n+2)(√(1+(n+3)(n+5))))) =(n+1)(√(1+(n+2)(√(1+(n+3)(√((n+5)^2 )))))) and so on putting n=l 8=2(√(1+3(√(1+4(√(1+5(√(1+.......)))))))) is the qiestion ok?

$$\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{3}\right)=\left(\mathrm{n}+\mathrm{1}\right)\sqrt{\left(\mathrm{n}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$=\left(\mathrm{n}+\mathrm{1}\right)\sqrt{\mathrm{1}+\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{4}\right)} \\ $$$$=\left(\mathrm{n}+\mathrm{1}\right)\sqrt{\mathrm{1}+\left(\mathrm{n}+\mathrm{2}\right)\sqrt{\left(\mathrm{n}+\mathrm{4}\right)^{\mathrm{2}} }} \\ $$$$=\left(\mathrm{n}+\mathrm{1}\right)\sqrt{\mathrm{1}+\left(\mathrm{n}+\mathrm{2}\right)\sqrt{\mathrm{1}+\left(\mathrm{n}+\mathrm{3}\right)\left(\mathrm{n}+\mathrm{5}\right)}} \\ $$$$=\left(\mathrm{n}+\mathrm{1}\right)\sqrt{\mathrm{1}+\left(\mathrm{n}+\mathrm{2}\right)\sqrt{\mathrm{1}+\left(\mathrm{n}+\mathrm{3}\right)\sqrt{\left(\mathrm{n}+\mathrm{5}\right)^{\mathrm{2}} }}} \\ $$$$\mathrm{and}\:\mathrm{so}\:\mathrm{on} \\ $$$$\mathrm{putting}\:\mathrm{n}=\mathrm{l} \\ $$$$\mathrm{8}=\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+\mathrm{5}\sqrt{\mathrm{1}+.......}}}} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{qiestion}\:\mathrm{ok}? \\ $$

Commented by ajfour last updated on 22/May/19

yes Sir, very fine one this is.

$$\mathrm{yes}\:\mathrm{Sir},\:\mathrm{very}\:\mathrm{fine}\:\mathrm{one}\:\mathrm{this}\:\mathrm{is}. \\ $$

Commented by Prithwish sen last updated on 22/May/19

thank you.

$$\mathrm{thank}\:\mathrm{you}. \\ $$

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