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Question Number 34186 by Joel578 last updated on 02/May/18

$$\mathrm{Let}{x}_1=0,x_2=1\mathrm{and}{x}_n=\frac{1}{2}(x_{n-1}+x_{n-2})$$$$\mathrm{Show}\mathrm{that}$$$$x_n=\frac{2^{n-1}+{(-1)}^n}{3.2^{n-2}}$$

Commented by abdo mathsup 649 cc last updated on 02/May/18

$$\iff 2x_n=x_{n-1}+x_{n-2}\iff 2x_{n+2}=x_{n+1}+x_n$$$$\iff 2x_{n+2}-x_{n+1}-x_n=0thecsrscteristicequation$$$$is2x^2-x-1=0$$$$\mathrm{\Delta }=1-4\left(2\right)(-1)=9>0$$$$x_1=\frac{1+3}{4}=1,x_2=\frac{1-3}{4}=-\frac{1}{2}therootsaresimples$$$$\Rightarrow x_n=\alpha 1^n+\beta {(-\frac{1}{2})}^n=\alpha +\beta {(-\frac{1}{2})}^n$$$$x_1=0\Rightarrow \alpha -\frac{\beta }{2}=0\Rightarrow \beta =2\alpha$$$$x_2=1\Rightarrow \alpha +\frac{\beta }{4}=1\Rightarrow 4\alpha +\beta =4\Rightarrow 4\alpha +2\alpha =4\Rightarrow$$$$6\alpha =4\Rightarrow \alpha =\frac{2}{3}and\beta =\frac{4}{3}\Rightarrow$$$$x_n=\frac{2}{3}+\frac{4}{3}{(-\frac{1}{2})}^n=\frac{2}{3}+\frac{4{(-1)}^n}{3.2^n}$$$$=\frac{1}{3}\left(\frac{2^{n+1}+4.{(-1)}^n}{2^n}\right)=\frac{4}{3}\left(\frac{2^{n-1}+{(-1)}^n}{2^n}\right)$$$$=\frac{2^{n-1}+{(-1)}^n}{3.2^{n-2}}.$$

Answered by candre last updated on 02/May/18

$$x_1=\frac{2^{1-1}+{(-1)}^1}{3\cdot 2^{1-2}}=\frac{1-1}{3/2}=0$$$$x_2=\frac{2^{2-1}+{(-1)}^2}{3\cdot 2^{2-2}}=\frac{2+1}{3}=1$$$$x_{n-1}=\frac{2^{n-2}+{(-1)}^{n-1}}{3\cdot 2^{n-3}}$$$$x_{n-2}=\frac{2^{n-3}+{(-1)}^{n-2}}{3\cdot 2^{n-4}}$$$$\frac{x_{n-1}+x_{n-2}}{2}=\frac{1}{2}(\frac{2^{n-2}+{(-1)}^{n-1}}{3\cdot 2^{n-3}}+\frac{2^{n-3}+{(-1)}^{n-2}}{3\cdot 2^{n-4}})$$$$=\frac{1}{2^{n+1}\cdot 3}(\frac{2^{n-2}-{(-1)}^n}{2^{-3}}+\frac{2^{n-3}+{(-1)}^n}{2^{-4}})$$$$=\frac{1}{2^{n+1}\cdot 3}[2^3(2^{n-2}-{(-1)}^n)+2^4(2^{n-3}+{(-1)}^n)]$$$$=\frac{2^3{2}^{n-2}-2^3{(-1)}^n+2^4{2}^{n-3}+2^4{(-1)}^n}{2^{n+1}\cdot 3}$$$$=\frac{2^{n+2}+(2^4-2^3){(-1)}^n}{2^{n+1}\cdot 3}$$$$=\frac{2^3(2^{n-1}+(2-1){(-1)}^n)}{2^{n+1}\cdot 3}$$$$=\frac{2^{n-1}+{(-1)}^n}{2^{n-2}\cdot 3}$$

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