Question Number 208135 by efronzo1 last updated on 06/Jun/24
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Answered by A5T last updated on 06/Jun/24
$${r}^{{n}+\mathrm{2}} ={r}^{{n}+\mathrm{1}} +\frac{{r}^{{n}} }{\mathrm{2}}\Rightarrow{r}^{\mathrm{2}} ={r}+\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{r}=\frac{\mathrm{1}\underset{−} {+}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${a}_{{n}} ={c}_{\mathrm{1}} \left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}} +{c}_{\mathrm{2}} \left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}} \\ $$$${a}_{\mathrm{1}} =\mathrm{1}\Rightarrow{c}_{\mathrm{1}} \left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)+{c}_{\mathrm{2}} \left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\mathrm{1} \\ $$$${a}_{\mathrm{2}} =\mathrm{2}\Rightarrow{c}_{\mathrm{1}} \left(\frac{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{4}}\right)+{c}_{\mathrm{2}} \left(\frac{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{4}}\right)=\mathrm{2} \\ $$$$\Rightarrow{c}_{\mathrm{1}} =\mathrm{1};{c}_{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow{a}_{{n}} =\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)^{{n}} +\left(\mathrm{1}−\sqrt{\mathrm{3}}\right)^{{n}} }{\mathrm{2}^{{n}} }\Rightarrow{a}_{\mathrm{11}} =\frac{\mathrm{989}}{\mathrm{32}} \\ $$