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

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

Commented by ajfour last updated on 22/May/19

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

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

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

Commented by ajfour last updated on 22/May/19

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

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

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

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