Question Number 175375 by Linton last updated on 28/Aug/22 | ||
Answered by mr W last updated on 28/Aug/22 | ||
$${g}_{{n}+\mathrm{1}} −\mathrm{3}{g}_{{n}} +\mathrm{2}{g}_{{n}−\mathrm{1}} =\mathrm{0} \\ $$$${r}^{\mathrm{2}} −\mathrm{3}{r}+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow{r}=\mathrm{1},\:\mathrm{2} \\ $$$$\Rightarrow{g}_{{n}} ={a}+{b}\mathrm{2}^{{n}} \\ $$ | ||
Commented by Tawa11 last updated on 28/Aug/22 | ||
$$\mathrm{Great}\:\mathrm{sir} \\ $$ | ||