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Question Number 175375 by Linton last updated on 28/Aug/22

Answered by mr W last updated on 28/Aug/22

g_(n+1) −3g_n +2g_(n−1) =0  r^2 −3r+2=0  ⇒r=1, 2  ⇒g_n =a+b2^n

$${g}_{{n}+\mathrm{1}} −\mathrm{3}{g}_{{n}} +\mathrm{2}{g}_{{n}−\mathrm{1}} =\mathrm{0} \\ $$$${r}^{\mathrm{2}} −\mathrm{3}{r}+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow{r}=\mathrm{1},\:\mathrm{2} \\ $$$$\Rightarrow{g}_{{n}} ={a}+{b}\mathrm{2}^{{n}} \\ $$

Commented by Tawa11 last updated on 28/Aug/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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