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Question Number 109950 by 675480065 last updated on 26/Aug/20

Answered by Aziztisffola last updated on 26/Aug/20

$$\left(\mathrm{i}\right){(\mathrm{x}-1)}^2>0\Rightarrow {\mathrm{x}}^2-2\mathrm{x}+1>0$$$$\Rightarrow {\mathrm{x}}^2>2\mathrm{x}-1\underset{2\mathrm{x}-1>1}{\stackrel{\mathrm{x}>1}{\Rightarrow }}\frac{{\mathrm{x}}^2}{2\mathrm{x}-1}>1$$$$\Rightarrow \mathrm{g}\left(\mathrm{x}\right)>1$$$$\left(\mathrm{ii}\right){\mathrm{w}}_{\mathrm{n}+1}=\ln \left({\mathrm{v}}_{\mathrm{n}+1}\right)=\ln (1-\frac{1}{{\mathrm{u}}_{\mathrm{n}+1}})$$$$=\ln (1-\frac{{2\mathrm{u}}_{\mathrm{n}}-1}{{\mathrm{u}}_{\mathrm{n}}^2})=\ln \left(\frac{{\mathrm{u}}_{\mathrm{n}}^2-{2\mathrm{u}}_{\mathrm{n}}+1}{{\mathrm{u}}_{\mathrm{n}}^2}\right)$$$$=\ln \left(\frac{{({\mathrm{u}}_{\mathrm{n}}-1)}^2}{{\mathrm{u}}_{\mathrm{n}}^2}\right)=\ln \left({\left(\frac{{\mathrm{u}}_{\mathrm{n}}-1}{{\mathrm{u}}_{\mathrm{n}}}\right)}^2\right)$$$$=2\ln \left(\frac{{\mathrm{u}}_{\mathrm{n}}-1}{{\mathrm{u}}_{\mathrm{n}}}\right)=2\ln \left({\mathrm{v}}_{\mathrm{n}}\right)={2\mathrm{w}}_{\mathrm{n}}$$$$\Rightarrow \mathrm{\forall }\mathrm{n}\in \mathbb{N}:{\mathrm{w}}_{\mathrm{n}+1}={2\mathrm{w}}_{\mathrm{n}}$$$$\Rightarrow \left({\mathrm{w}}_{\mathrm{n}}\right)\mathrm{is}\mathrm{geometric}\mathrm{sequence}\mathrm{with}$$$$\mathrm{ratio}\mathrm{q}=2\mathrm{and}\mathrm{first}\mathrm{term}{\mathrm{w}}_0=\ln \left({\mathrm{v}}_0\right)$$$${\mathrm{w}}_0=\ln \left(\frac{{\mathrm{u}}_0-1}{{\mathrm{u}}_0}\right)=\ln \left(\frac{1}{2}\right)=-\ln 2$$

Answered by Aziztisffola last updated on 26/Aug/20

$$\left(\mathrm{iii}\right){\mathrm{w}}_{\mathrm{n}}=-\ln \left(2\right)\times 2^{\mathrm{n}}=\ln \left({\mathrm{v}}_{\mathrm{n}}\right)$$$$\Rightarrow {\mathrm{v}}_{\mathrm{n}}={\mathrm{e}}^{-\ln 2\times 2^{\mathrm{n}}}={\left({\mathrm{e}}^{\ln 2}\right)}^{-2^{\mathrm{n}}}=2^{-2^{\mathrm{n}}}$$

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