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Question Number 31214 by Tinkutara last updated on 03/Mar/18

Commented by abdo imad last updated on 03/Mar/18

$$wehave{u}_{n+1}=3u_n-2u_{n-1}\Rightarrow u_{n+2}-3u_{n+1}+2u_n=0$$$$thecaracteristicequationis{r}^2-3r+2=0roots?$$$$\mathrm{\Delta }=9-8=1\Rightarrow r_1=\frac{3+1}{2}=2and{r}_2=\frac{3-1}{2}=1sotheroots$$$$aresimple\Rightarrow u_n={ar}_1^n+{br}_2^n=1+a{2}^{n}wehave$$$$u_0=1+a=2\Rightarrow a=1andweverifythst{u}_1=3\Rightarrow$$$$u_n=1+2^n.$$

Commented by Tinkutara last updated on 04/Mar/18

Thanks Sir!

Answered by math solver last updated on 03/Mar/18

$$\left(2\right)2^{\mathrm{n}}+1.$$

Answered by ajfour last updated on 03/Mar/18

$$u_{n+1}=3(3u_{n-1}-2u_{n-2})-2u_{n-1}$$$$=7u_{n-1}-6u_{n-2}$$$$=7(3u_{n-2}-2u_{n-3})-6u_{n-2}$$$$=15u_{n-2}-14u_{n-3}$$$$=15(3u_{n-3}-2u_{n-4})-14u_{n-3}$$$$=31u_{n-3}-30u_{n-4}$$$$=(2^5-1)u_{n-3}-(2^5-2)u_{n-4}$$$$=(2^{n+1}-1)u_{n-(n-1)}-(2^{n+1}-2)u_{n-n}$$$$=(2^{n+1}-1)\times 3-(2^{n+1}-2)\times 2$$$$=2^{n+1}+1$$$$\Rightarrow {\mathit{u}}_{\mathit{n}}=2^{\mathit{n}}+1.$$

Commented by Tinkutara last updated on 03/Mar/18

Thanks for it!

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