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Question Number 145944 by alcohol last updated on 09/Jul/21

Commented by alcohol last updated on 10/Jul/21

$$nobody?$$

Answered by mathmax by abdo last updated on 10/Jul/21

$$1)\{\begin{array}{l}{\mathrm{u}}_0=-6\\ {\mathrm{u}}_{\mathrm{n}+1}=\frac{1}{3}({\mathrm{u}}_{\mathrm{n}}-{2\mathrm{v}}_{\mathrm{n}})\mathrm{and}\{\begin{array}{l}{\mathrm{v}}_0=1\\ {\mathrm{v}}_{\mathrm{n}+1}=\frac{1}{2}({\mathrm{u}}_{\mathrm{n}}+{3\mathrm{v}}_{\mathrm{n}})\end{array}\end{array}$$$${\mathrm{w}}_{\mathrm{n}}={\mathrm{u}}_{\mathrm{n}}+{\mathrm{v}}_{\mathrm{n}}$$$$\frac{{\mathrm{w}}_{\mathrm{n}+1}}{{\mathrm{w}}_{\mathrm{n}}}=\frac{{\mathrm{u}}_{\mathrm{n}+1}+{\mathrm{v}}_{\mathrm{n}+1}}{{\mathrm{u}}_{\mathrm{n}}+{\mathrm{v}}_{\mathrm{n}}}=\frac{\frac{1}{3}({\mathrm{u}}_{\mathrm{n}}-{2\mathrm{v}}_{\mathrm{n}})+\frac{1}{2}({\mathrm{u}}_{\mathrm{n}}+{3\mathrm{v}}_{\mathrm{n}})}{{\mathrm{u}}_{\mathrm{n}}+{\mathrm{v}}_{\mathrm{n}}}$$$$=\frac{\frac{5}{6}{\mathrm{u}}_{\mathrm{n}}+(-\frac{2}{3}+\frac{3}{2}){\mathrm{v}}_{\mathrm{n}}}{{\mathrm{u}}_{\mathrm{n}}+{\mathrm{v}}_{\mathrm{n}}}=\frac{5}{6}\times \frac{{\mathrm{u}}_{\mathrm{n}}+{\mathrm{v}}_{\mathrm{n}}}{{\mathrm{u}}_{\mathrm{n}}+{\mathrm{v}}_{\mathrm{n}}}=\frac{5}{6}\Rightarrow$$$${\mathrm{w}}_{\mathrm{n}}\mathrm{is}\mathrm{geometric}\mathrm{with}\mathrm{q}=\frac{5}{6}$$$${\mathrm{w}}_{\mathrm{n}}={\mathrm{w}}_0{\mathrm{q}}^{\mathrm{n}}\mathrm{and}\mathrm{w}0={\mathrm{u}}_0+{\mathrm{v}}_0=-5\Rightarrow {\mathrm{w}}_{\mathrm{n}}=-5{\left(\frac{5}{6}\right)}^{\mathrm{n}}$$$$2)\mathrm{we}\mathrm{have}\mid \frac{5}{6}\mid <1\Rightarrow {\lim }_{\mathrm{n}\to \mathrm{\infty }}{\mathrm{w}}_{\mathrm{n}}=0$$

Commented by physicstutes last updated on 11/Jul/21

$$\left(3\right)u_{n+1}-u_n=\frac{1}{3}(u_n-2v_n)-u_n$$$$=\frac{1}{3}{u}_n-\frac{2}{3}{v}_n-u_n$$$$=-\frac{1}{3}(u_n+v_n)=-\frac{1}{3}(-5{\left(\frac{5}{6}\right)}^n)=\frac{5}{3}{\left(\frac{5}{6}\right)}^n>0$$$$\mathrm{Hence}{u}_n\mathrm{is}\mathrm{increasing}.$$$$$$

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