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Question Number 142987 by mathmax by abdo last updated on 08/Jun/21

$$\mathrm{find}\mathrm{the}\mathrm{sequence}{\mathrm{u}}_{\mathrm{n}}\mathrm{wich}\mathrm{verify}{\mathrm{u}}_{\mathrm{n}+1}={\mathrm{u}}_{\mathrm{n}}-\lambda {\mathrm{u}}_{\mathrm{n}-1}$$$$\lambda \mathrm{real}$$

Answered by Dwaipayan Shikari last updated on 08/Jun/21

$$u_n={mr}^n$$$${mr}^{n+1}={mr}^n-\lambda {mr}^{n-1}$$$$\Rightarrow r^2-r+\lambda =0\Rightarrow r=\frac{1\pm \sqrt{1-4\lambda }}{2}\lambda ⩽\frac{1}{4}r\in \mathbb{R}$$$$ifr\in \mathbb{C}then-\mathrm{\infty }<\lambda <\mathrm{\infty }$$$$u_n=\mathrm{\Lambda }{\left(\frac{1+\sqrt{1-4\lambda }}{2}\right)}^n-\mathrm{\Phi }{\left(\frac{1-\sqrt{1-4\lambda }}{2}\right)}^n$$$$u_0=\mathrm{\Lambda }-\mathrm{\Phi }$$$$u_1=\frac{\mathrm{\Lambda }-\mathrm{\Phi }}{2}+\frac{\mathrm{\Lambda }+\mathrm{\Phi }}{2}\sqrt{1-4\lambda }\Rightarrow \frac{1}{\sqrt{1-4\lambda }}(2u_1-u_0)=\mathrm{\Lambda }+\mathrm{\Phi }$$$$\mathrm{\Lambda }=\frac{1}{2}.\frac{(2u_1-u_0)}{\sqrt{1-4\lambda }}+\frac{u_0}{2}$$$$\mathrm{\Phi }=\frac{1}{2}.\frac{(2u_1-u_0)}{\sqrt{1-4\lambda }}-\frac{u_0}{2}$$$$u_n=\frac{(2u_1-u_0)}{2\sqrt{1-4\lambda }}\{{\left(\frac{1+\sqrt{1-4\lambda }}{2}\right)}^n-{\left(\frac{1-\sqrt{1-4\lambda }}{2}\right)}^n\}+\frac{u_0}{2}\{{\left(\frac{1+\sqrt{1-4\lambda }}{2}\right)}^n-{\left(\frac{1-\sqrt{1-4\lambda }}{2}\right)}^n\}$$

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