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Question Number 105205 by Ar Brandon last updated on 26/Jul/20

Commented by Aziztisffola last updated on 26/Jul/20

$$\mathrm{ce}{\mathrm{n}}^{\prime }\mathrm{est}\mathrm{pas}\mathrm{tres}\mathrm{lisible}.$$

Commented by Ar Brandon last updated on 26/Jul/20

D'accord, désolé pour l'inconvenance.

Answered by mathmax by abdo last updated on 27/Jul/20

$${\mathrm{u}}_{\mathrm{n}}=\frac{3}{4}{\mathrm{u}}_{\mathrm{n}-1}+\frac{1}{4}{\mathrm{u}}_{\mathrm{n}-2}\Rightarrow {4\mathrm{u}}_{\mathrm{n}}={3\mathrm{u}}_{\mathrm{n}-1}+{\mathrm{u}}_{\mathrm{n}-2}\Rightarrow {4\mathrm{u}}_{\mathrm{n}+2}={3\mathrm{u}}_{\mathrm{n}+1}+{\mathrm{u}}_{\mathrm{n}}\Rightarrow$$$${4\mathrm{u}}_{\mathrm{n}+2}-{3\mathrm{u}}_{\mathrm{n}+1}-{\mathrm{u}}_{\mathrm{n}}=0\to {4\mathrm{r}}^2-3\mathrm{r}-1=0$$$$\mathrm{\Delta }=9-4(-4)=25\Rightarrow {\mathrm{r}}_1=\frac{3+5}{8}=1\mathrm{and}{\mathrm{r}}_2=\frac{3-5}{8}=-\frac{1}{4}\Rightarrow$$$${\mathrm{u}}_{\mathrm{n}}=\alpha +\beta {(-\frac{1}{4})}^{\mathrm{n}}$$$${\mathrm{u}}_0=\alpha +\beta$$$${\mathrm{u}}_1=\alpha -\frac{\beta }{4}\Rightarrow \frac{5}{4}\beta ={\mathrm{u}}_0-{\mathrm{u}}_1\Rightarrow \beta =\frac{4}{5}({\mathrm{u}}_0-{\mathrm{u}}_1)$$$$\alpha ={\mathrm{u}}_0-\beta ={\mathrm{u}}_0-\frac{4}{5}({\mathrm{u}}_{\mathrm{o}}-{\mathrm{u}}_1)=\frac{1}{5}{\mathrm{u}}_0+\frac{4}{5}{\mathrm{u}}_1$$$$\Rightarrow {\mathrm{u}}_{\mathrm{n}}=\frac{1}{5}({\mathrm{u}}_0+{4\mathrm{u}}_1)+\frac{4}{5}({\mathrm{u}}_0-{\mathrm{u}}_1){(-\frac{1}{4})}^{\mathrm{n}}$$$$\mathrm{if}{\mathrm{u}}_0={\mathrm{u}}_1\Rightarrow {\mathrm{u}}_{\mathrm{n}}={\mathrm{u}}_0\mathrm{\forall }\mathrm{n}\mathrm{so}{\mathrm{u}}_{\mathrm{n}}\mathrm{is}\mathrm{constante}$$$$2){\mathrm{v}}_{\mathrm{n}}={\mathrm{u}}_{\mathrm{n}}-{\mathrm{u}}_{\mathrm{n}-1}\Rightarrow {\mathrm{v}}_{\mathrm{n}+1}={\mathrm{u}}_{\mathrm{n}+1}-{\mathrm{u}}_{\mathrm{n}}=\frac{3}{4}{\mathrm{u}}_{\mathrm{n}}+\frac{1}{4}{\mathrm{u}}_{\mathrm{n}-1}-{\mathrm{u}}_{\mathrm{n}-1}$$$$=\frac{3}{4}{\mathrm{u}}_{\mathrm{n}}-\frac{3}{4}{\mathrm{u}}_{\mathrm{n}-1}=\frac{3}{4}({\mathrm{u}}_{\mathrm{n}}-{\mathrm{u}}_{\mathrm{n}-1})=\frac{3}{4}{\mathrm{v}}_{\mathrm{n}}\Rightarrow {\mathrm{v}}_{\mathrm{n}}\mathrm{is}\mathrm{geometric}\mathrm{with}$$$$\mathrm{q}=\frac{3}{4}\Rightarrow {\mathrm{v}}_{\mathrm{n}}={\mathrm{v}}_1\times {\left(\frac{3}{4}\right)}^{\mathrm{n}-1}$$$${\mathrm{v}}_1={\mathrm{u}}_1-{\mathrm{u}}_0\ne 0\Rightarrow {\mathrm{v}}_{\mathrm{n}}=({\mathrm{u}}_1-{\mathrm{u}}_0){\left(\frac{3}{4}\right)}^{\mathrm{n}-1}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{v}}_{\mathrm{k}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}({\mathrm{u}}_{\mathrm{k}}-{\mathrm{u}}_{\mathrm{k}-1})={\mathrm{u}}_1-{\mathrm{u}}_0+{\mathrm{u}}_2-{\mathrm{u}}_1+...+{\mathrm{u}}_{\mathrm{n}}-{\mathrm{u}}_{\mathrm{n}-1}$$$$={\mathrm{u}}_{\mathrm{n}}-{\mathrm{u}}_{\mathrm{o}}\Rightarrow {\mathrm{u}}_{\mathrm{n}}-{\mathrm{u}}_0=({\mathrm{u}}_1-{\mathrm{u}}_0)\sum _{\mathrm{k}=1}^{\mathrm{n}}{\left(\frac{3}{4}\right)}^{\mathrm{k}-1}(\mathrm{k}-1=\mathrm{p})$$$$=({\mathrm{u}}_1-{\mathrm{u}}_0)\sum _{\mathrm{k}=0}^{\mathrm{n}-1}{\left(\frac{3}{4}\right)}^{\mathrm{p}}=({\mathrm{u}}_1-{\mathrm{u}}_0)\times \frac{1-{\left(\frac{3}{4}\right)}^{\mathrm{n}}}{1-\frac{3}{4}}$$$$=4({\mathrm{u}}_1-{\mathrm{u}}_0)\times (1-{\left(\frac{3}{4}\right)}^{\mathrm{n}})\Rightarrow {\mathrm{u}}_{\mathrm{n}}={\mathrm{u}}_0+4({\mathrm{u}}_1-{\mathrm{u}}_0)(1-{\left(\frac{3}{4}\right)}^{\mathrm{n}})$$

Commented by Ar Brandon last updated on 27/Jul/20

Thanks Sir. Always willing to help ! ��

Commented by abdomathmax last updated on 27/Jul/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}\mathrm{sir}.$$

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