Question Number 168233 by alcohol last updated on 06/Apr/22
$$\begin{cases}{{u}_{\mathrm{0}} \:=\:\mathrm{3}\::\:{u}_{\mathrm{1}} \:=\:\mathrm{4}}\\{{u}_{{n}+\mathrm{1}} \:=\:{u}_{{n}} \:+\:\mathrm{6}{u}_{{n}−\mathrm{1}} }\end{cases} \\ $$$${Express}\:{u}_{{n}} \:{in}\:{terms}\:{of}\:{n} \\ $$
Answered by mr W last updated on 06/Apr/22
$${r}^{\mathrm{2}} −{r}−\mathrm{6}=\mathrm{0} \\ $$$$\left({r}+\mathrm{2}\right)\left({r}−\mathrm{3}\right)=\mathrm{0} \\ $$$${r}_{\mathrm{1}} =−\mathrm{2},\:{r}_{\mathrm{2}} =\mathrm{3} \\ $$$${u}_{{n}} ={A}×\left(−\mathrm{2}\right)^{{n}} +{B}×\mathrm{3}^{{n}} \\ $$$${u}_{\mathrm{0}} ={A}+{B}=\mathrm{3} \\ $$$${u}_{\mathrm{1}} =−\mathrm{2}{A}+\mathrm{3}{B}=\mathrm{4} \\ $$$$\Rightarrow{B}=\mathrm{2},\:{A}=\mathrm{1} \\ $$$$\Rightarrow{u}_{{n}} =\left(−\mathrm{2}\right)^{{n}} +\mathrm{2}×\mathrm{3}^{{n}} \\ $$