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2-1-3-1-5-1-7-1-11-1-13-1-17-x-x-




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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