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Question Number 175211 by MathsFan last updated on 23/Aug/22

$$\mathbf{Assume}\mathbf{that}\mathbf{the}\mathbf{sequence}\mathbf{terms}\mathbf{tend}$$$$\mathbf{to}\mathbf{the}\mathbf{constant}\mathbf{value}\mathit{u},\mathbf{so}\mathbf{that}\mathbf{as}$$$$\mathbf{n}\to \mathrm{\infty },{\mathit{u}}_{\mathit{n}-1}\to \mathit{u}\mathbf{and}{\mathit{u}}_{\mathit{n}}\to \mathit{u}.$$$$\left(\mathbf{i}\right)\mathbf{show}\mathbf{that}{\mathit{u}}^2+\mathit{u}-1=0$$$$\left(\mathbf{ii}\right)\mathbf{show}\mathbf{that}\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+.....}}}}=\frac{-1+\sqrt{5}}{2}$$

Answered by Rasheed.Sindhi last updated on 23/Aug/22

$$\left(\mathbf{ii}\right)\mathbf{u}=\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+.....}}}};\mathbf{u}>0$$$$\mathbf{u}=\frac{1}{1+\mathbf{u}}$$$${\mathbf{u}}^2+\mathbf{u}-1=0$$$$\mathbf{u}=\frac{-1+\sqrt{1-4\left(1\right)(-1)}}{2}=\frac{-1+\sqrt{5}}{2}$$$$[\frac{-1-\sqrt{5}}{2}<0]$$

Commented by MathsFan last updated on 24/Aug/22

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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