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Question Number 148502 by mathmax by abdo last updated on 28/Jul/21

$$\mathrm{let}\alpha \mathrm{and}\beta \mathrm{roots}\mathrm{of}{\mathrm{x}}^2+\mathrm{x}+2$$$$\mathrm{simplify}\sum _{\mathrm{k}=0}^{\mathrm{n}-1}({\alpha }^{\mathrm{k}}+{\beta }^{\mathrm{k}})\mathrm{and}\sum _{\mathrm{k}=0}^{\mathrm{n}-1}(\frac{1}{{\alpha }^{\mathrm{k}}}+\frac{1}{{\beta }^{\mathrm{k}}})$$

Answered by ArielVyny last updated on 28/Jul/21

$$x^2+x+2=0\to {(x+\frac{1}{2})}^2-\frac{1}{4}+\frac{8}{4}=0$$$$\to {(x+\frac{1}{2})}^2-{\left(i\frac{\sqrt{7}}{2}\right)}^2=0\to (x+\frac{1}{2}+i\frac{\sqrt{7}}{2})(x+\frac{1}{2}-i\frac{\sqrt{7}}{2})=0$$$$\alpha =-\frac{1+i\sqrt{7}}{2}and\beta =\frac{-1+i\sqrt{7}}{2}$$$$\mid \alpha \mid =\mid \beta \mid \sqrt{{(-\frac{1}{2})}^2+{\left(\frac{\sqrt{7}}{2}\right)}^2}=\sqrt{\frac{1}{4}+\frac{7}{4}}=\frac{2\sqrt{2}}{2}=\sqrt{2}$$$$\alpha =\sqrt{2}(-\frac{1}{2\sqrt{2}}-i\frac{\sqrt{7}}{2\sqrt{2}})=\sqrt{2}(-\frac{\sqrt{2}}{4}-i\frac{\sqrt{14}}{4})=\sqrt{2}{e}^{i\theta }$$$$\beta =\sqrt{2}(-\frac{\sqrt{2}}{4}+\frac{\sqrt{14}}{4})=\sqrt{2}{e}^{-i\theta }$$$$\theta =110,7048111\sim$$$$U_n=\sum _{k=0}^{n-1}({\alpha }^k+{\beta }^k)=\underset{k=0}{\sum ^{n-1}}{\alpha }^k+\underset{k=0}{\sum ^{n-1}}{\beta }^k$$$$\underset{k=0}{\sum ^{n-1}}{\left(e^{i\theta }\right)}^k+\underset{k=0}{\sum ^{n-1}}{\left(e^{-i\theta }\right)}^k$$$$\frac{1-{\left(e^{i\theta }\right)}^n}{1-e^{i\theta }}+\frac{1-{\left(e^{-i\theta }\right)}^n}{1-e^{-i\theta }}=\frac{1-e^{in\theta }}{1-e^{i\theta }}+\frac{1-\frac{1}{e^{in\theta }}}{1-\frac{1}{e^{i\theta }}}$$$$=\frac{1-e^{in\theta }}{1-e^{i\theta }}+\frac{\frac{e^{in\theta }-1}{e^{in\theta }}}{\frac{e^{i\theta }-1}{e^{i\theta }}}=\frac{1-e^{in\theta }}{1-e^{i\theta }}-\frac{e^{i\theta (n+1)}-e^{i\theta }}{e^{in\theta }}\frac{1}{1-e^{i\theta }}$$$$U_n=\frac{1}{1-e^{i\theta }}(1-e^{in\theta }-\frac{e^{i\theta (n+1)}-e^{i\theta }}{e^{in\theta }})$$$$U_n=\frac{1}{1-e^{i\theta }}(1-e^{in\theta }-e^{i\theta }-e^{i\theta (1-n)})$$$$U_n=\frac{1}{1-e^{i\theta }}(1-e^{in\theta }-e^{i\theta }(1+e^{-in}))$$

Answered by Olaf_Thorendsen last updated on 28/Jul/21

$$\mathrm{Let}{u}_0=2,u_1=-1$$$$\mathrm{and}\mathrm{\forall }n⩾2,u_n=-u_{n-1}-2u_{n-2}$$$$\mathrm{We}\mathrm{solve}{r}^2+r+2=0.$$$$\mathrm{The}\mathrm{roots}\mathrm{are}\mathrm{noted}\alpha \mathrm{and}\beta .$$$$\mathrm{Then}{u}_n=\lambda {\alpha }^n+\mu {\beta }^n.$$$$\mathrm{With}{u}_0\mathrm{and}{u}_1,\lambda =\mu =1.$$$$\Rightarrow \mathrm{\forall }n\in \mathbb{N},u_n={\alpha }^n+{\beta }^n$$$$$$$${\mathrm{S}}_{n+2}=\underset{k=0}{\sum ^{n+2}}({\alpha }^k+{\beta }^k)=\underset{k=0}{\sum ^{n+2}}{u}_k$$$${\mathrm{S}}_{n+2}=u_0+u_1+\underset{k=2}{\sum ^{n+2}}{u}_k$$$${\mathrm{S}}_{n+2}=u_0+u_1+\underset{k=0}{\sum ^n}{u}_{k+2}$$$${\mathrm{S}}_{n+2}=u_0+u_1-\underset{k=0}{\sum ^n}{u}_{k+1}-2\underset{k=0}{\sum ^n}{u}_k$$$${\mathrm{S}}_{n+2}=u_0+u_1-({\mathrm{S}}_{n+2}-u_0-u_{n+2})$$$$-2({\mathrm{S}}_{n+2}-u_{n+1}-u_{n+2})$$$$$$$${4\mathrm{S}}_{n+2}=3u_{n+2}+2u_{n+1}+2u_0+u_1$$$${\mathrm{S}}_{n+2}=\frac{3}{4}({\alpha }^{n+2}+{\beta }^{n+2})+\frac{1}{2}({\alpha }^{n+1}+{\beta }^{n+1})+\frac{3}{4}$$$$$$$$\underset{k=0}{\sum ^{n-1}}({\alpha }^k+{\alpha }^k)={\mathrm{S}}_{n-1}$$$$=\frac{3}{4}({\alpha }^{n-1}+{\beta }^{n-1})+\frac{1}{2}({\alpha }^{n-2}+{\beta }^{n-2})+\frac{3}{4}$$$$=\frac{1}{4}{\alpha }^{n-2}(3\alpha +2)+\frac{1}{4}{\beta }^{n-2}(3\beta +2)+\frac{3}{4}$$$$=\frac{1}{4}{\alpha }^{n-2}(2\alpha -{\alpha }^2)+\frac{1}{4}{\beta }^{n-2}(2\beta -{\beta }^2)+\frac{3}{4}$$$$$$$$\underset{k=0}{\sum ^{n-1}}({\alpha }^k+{\alpha }^k)=\frac{1}{4}{\alpha }^{n-1}(2-\alpha )+\frac{1}{4}{\beta }^{n-1}(2-\beta )+\frac{3}{4}$$$$$$$$$$$$\alpha +\beta =-\frac{b}{a}=-1\mathrm{and}\alpha \beta =\frac{c}{a}=2$$$$(x-\frac{1}{\alpha })(x-\frac{1}{\beta })=x^2-(\frac{1}{\alpha }+\frac{1}{\beta })x+\frac{1}{\alpha \beta }$$$$=x^2-\frac{\alpha +\beta }{\alpha \beta }x+\frac{1}{\alpha \beta }$$$$=x^2+\frac{1}{2}x+\frac{1}{2}$$$$\Rightarrow \mathrm{For}\underset{k=0}{\sum ^{n-1}}(\frac{1}{{\alpha }^k}+\frac{1}{{\beta }^k}):\mathrm{same}\mathrm{kind}\mathrm{of}\mathrm{calculous}$$$$\mathrm{with}{u}_n=-\frac{1}{2}{u}_{n-1}-\frac{1}{2}{u}_{n-2}$$

Answered by mathmax by abdo last updated on 28/Jul/21

$${\mathrm{x}}^2+\mathrm{x}+2=0\to \mathrm{\Delta }=1-8=-7\Rightarrow {\mathrm{x}}_1=\frac{-1+\mathrm{i}\sqrt{7}}{2}\mathrm{and}{\mathrm{x}}_2=\frac{-1-\mathrm{i}\sqrt{7}}{2}$$$$\mid {\mathrm{x}}_1\mid =\frac{1}{2}\sqrt{1+7}=\frac{2\sqrt{2}}{2}\Rightarrow {\mathrm{x}}_1=\sqrt{2}(\frac{-1}{2\sqrt{2}}+\frac{\mathrm{i}\sqrt{7}}{2\sqrt{2}})=\sqrt{2}{\mathrm{e}}^{-\mathrm{iarctan}\left(\sqrt{7}\right)}$$$$\Rightarrow {\mathrm{x}}_2=\sqrt{2}{\mathrm{e}}^{\mathrm{iarctan}\left(\sqrt{7}\right)}\Rightarrow$$$${\mathrm{A}}_{\mathrm{n}}=\sum _{\mathrm{k}=0}^{\mathrm{n}-1}{\alpha }^{\mathrm{k}}+{\beta }^{\mathrm{k}}=\sum _{\mathrm{k}=0}^{\mathrm{n}-1}{\left(\sqrt{2}{\mathrm{e}}^{\mathrm{iarctan}\left(\sqrt{7}\right)}\right)}^{\mathrm{k}}+{\left(\sqrt{2}{\mathrm{e}}^{-\mathrm{iarctan}\left(\sqrt{7}\right)}\right)}^{\mathrm{k}}$$$$=\sum _{\mathrm{k}=0}^{\mathrm{n}-1}{\left(\sqrt{2}\right)}^{\mathrm{k}}\{{\mathrm{e}}^{\mathrm{ikarctan}\left(\sqrt{7}\right)}+{\mathrm{e}}^{-\mathrm{ikarctan}\left(\sqrt{7}\right)}\}$$$$=2\sum _{\mathrm{k}=0}^{\mathrm{n}-1}{\left(\sqrt{2}\right)}^{\mathrm{k}}\cos \left(\mathrm{karctan}\left(\sqrt{7}\right)\right)$$$$\mathrm{let}{\mathrm{T}}_{\mathrm{k}}\left(\mathrm{x}\right)=\cos \left(\mathrm{karctanx}\right)\Rightarrow {\mathrm{A}}_{\mathrm{n}}=2\sum _{\mathrm{k}=0}^{\mathrm{n}-1}{\left(\sqrt{2}\right)}^{\mathrm{k}}{\mathrm{T}}_{\mathrm{k}}\left(\sqrt{7}\right)$$$$\sum _{\mathrm{k}=0}^{\mathrm{n}-1}(\frac{1}{{\alpha }^{\mathrm{k}}}+\frac{1}{{\beta }^{\mathrm{k}}})=\sum _{\mathrm{k}=0}^{\mathrm{n}-1}\left(\frac{{\alpha }^{\mathrm{k}}+{\beta }^{\mathrm{k}}}{{\left(\alpha \beta \right)}^{\mathrm{k}}}\right)$$$$=\sum _{\mathrm{k}=0}^{\mathrm{n}-1}\frac{1}{2^{\mathrm{k}}}({\alpha }^{\mathrm{k}}+{\beta }^{\mathrm{k}})=2\sum _{\mathrm{k}=0}^{\mathrm{n}-1}\frac{1}{2^{\mathrm{k}}}{\left(\sqrt{2}\right)}^{\mathrm{k}}\cos \left(\mathrm{karctan}\left(\sqrt{7}\right)\right)$$$$=2\sum _{\mathrm{k}=0}^{\mathrm{n}-1}{\left(\sqrt{2}\right)}^{-\mathrm{k}}{\mathrm{T}}_{\mathrm{k}}\left(\sqrt{7}\right)$$$$$$

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