let-a-1-gt-a-2-gt-0-and-a-n-1-a-n-a-n-1-where-n-is-greater-than-equal-to-2-Then-The-sequence-a-2n-is-1-monotonic-increasing-2-monotonic-decreasing-3-non-monotonic-4-unbound

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Question Number 16110 by vpawarksp@gmail.com last updated on 18/Jun/17

$$\mathrm{let}{\mathrm{a}}_1>{\mathrm{a}}_2>0\mathrm{and}{\mathrm{a}}_{\mathrm{n}+1}=\sqrt{{\mathrm{a}}_{\mathrm{n}}{\mathrm{a}}_{\mathrm{n}-1}}$$$$\mathrm{where}\mathrm{n}\mathrm{is}\mathrm{greater}\mathrm{than}\mathrm{equal}\mathrm{to}2$$$$\mathrm{Then}$$$$\mathrm{The}\mathrm{sequence}\left\{{\mathrm{a}}_{2\mathrm{n}}\right\}\mathrm{is}$$$$\left(1\right)\mathrm{monotonic}\mathrm{increasing}$$$$\left(2\right)\mathrm{monotonic}\mathrm{decreasing}$$$$\left(3\right)\mathrm{non}\mathrm{monotonic}$$$$\left(4\right)\mathrm{unbounded}$$$$$$

Commented by RasheedSoomro last updated on 18/Jun/17

$$\mathrm{let}{\mathrm{a}}_1>{\mathrm{a}}_2>0\mathrm{and}{\mathrm{a}}_{\mathrm{n}+1}=\sqrt{{\mathrm{a}}_{\mathrm{n}}{\mathrm{a}}_{\mathrm{n}-1}}$$$${\mathrm{a}}_{\mathrm{n}}=\frac{{\left({\mathrm{a}}_{\mathrm{n}+1}\right)}^2}{{\mathrm{a}}_{\mathrm{n}-1}}\dots \dots \dots \dots \dots \left(\mathrm{i}\right)$$$$\mathrm{n}\to \mathrm{n}-1\Rightarrow {\mathrm{a}}_{\mathrm{n}}=\sqrt{{\mathrm{a}}_{\mathrm{n}-1}{\mathrm{a}}_{\mathrm{n}-2}}\dots \dots \dots \left(\mathrm{ii}\right)$$$$\left(\mathrm{i}\right)\mathrm{\&}\left(\mathrm{ii}\right)$$$$\frac{{\left({\mathrm{a}}_{\mathrm{n}+1}\right)}^2}{{\mathrm{a}}_{\mathrm{n}-1}}=\sqrt{{\mathrm{a}}_{\mathrm{n}-1}{\mathrm{a}}_{\mathrm{n}-2}}$$$$$$

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