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Question Number 146902 by mathmax by abdo last updated on 16/Jul/21
$$\mathrm{let}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{roots}\:\mathrm{of}\:\:\mathrm{z}^{\mathrm{2}} +\mathrm{3z}+\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{simlify}\:\mathrm{U}_{\mathrm{n}} =\:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\left(\alpha^{\mathrm{k}} \:+\beta^{\mathrm{k}} \right) \\ $$$$\mathrm{and}\:\mathrm{V}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\left(\frac{\mathrm{1}}{\alpha^{\mathrm{k}} }+\frac{\mathrm{1}}{\beta^{\mathrm{k}} }\right) \\ $$
Answered by ArielVyny last updated on 18/Jul/21
![z^2 +3z+5=0→(z+(3/2))^2 −(9/4)+((20)/4)=0 (z+(3/2))^2 +((11)/4)=0 (z+(3/2)−i((√(11))/2))(z+(3/2)+i((√(11))/2))=0 then α=−(3/2)+i((√(11))/2) and β=−(3/2)−i((√(11))/2) U_n =Σ_(k=0) ^n (α^k +β^k ) ∣α∣=∣−(3/2)+i((√(11))/2)∣=(√((−(3/2))^2 +(((√(11))/2))^2 )) ∣α∣=(√((9/4)+((11)/4)))=((√(20))/2) α=((√(20))/2)(−((3/2)/((√(20))/2))+i(((√(11))/2)/((√(20))/2)))=((√(20))/2)(−(3/( (√(20))))+i((√(11))/( (√(20))))) α=((√(20))/2)e^(iθ) tel que { ((cosθ=−(3/( (√(20)))))),((sinθ=(√((11)/(20))))) :} β=((√(20))/2)e^(−iθ) U_n =Σ_(k=0) ^n [(((√(20))/2))^k e^(ikθ) +(((√(20))/2))^k e^(−ikθ) ] U_n =((1−(((√(20))/2)e^(iθ) )^(n+1) )/(1−(((√(20))/2)e^(iθ) )))+((1−(((√(20))/2)e^(−iθ) )^(n+1) )/(1−(((√(20))/2)e^(−iθ) ))) U_n =((1−((√(20))/2)e^(−iθ) −(((√(20))/2)e^(iθ) )^(n+1) +(((√(20))/2))^(n+2) e^(iθn) )/(1−(((√(20))/2)e^(−iθ) )−(((√(20))/2)e^(iθ) )+((20)/4))) to be continued....](Q147189.png)
$${z}^{\mathrm{2}} +\mathrm{3}{z}+\mathrm{5}=\mathrm{0}\rightarrow\left({z}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{20}}{\mathrm{4}}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({z}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{11}}{\mathrm{4}}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({z}+\frac{\mathrm{3}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\right)\left({z}+\frac{\mathrm{3}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$${then}\:\alpha=−\frac{\mathrm{3}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\:\:{and}\:\beta=−\frac{\mathrm{3}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{11}}}{\mathrm{2}} \\ $$$${U}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \left(\alpha^{{k}} +\beta^{{k}} \right) \\ $$$$\mid\alpha\mid=\mid−\frac{\mathrm{3}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\mid=\sqrt{\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\mid\alpha\mid=\sqrt{\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{11}}{\mathrm{4}}}=\frac{\sqrt{\mathrm{20}}}{\mathrm{2}} \\ $$$$\alpha=\frac{\sqrt{\mathrm{20}}}{\mathrm{2}}\left(−\frac{\frac{\mathrm{3}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{20}}}{\mathrm{2}}}+{i}\frac{\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{20}}}{\mathrm{2}}}\right)=\frac{\sqrt{\mathrm{20}}}{\mathrm{2}}\left(−\frac{\mathrm{3}}{\:\sqrt{\mathrm{20}}}+{i}\frac{\sqrt{\mathrm{11}}}{\:\sqrt{\mathrm{20}}}\right) \\ $$$$\alpha=\frac{\sqrt{\mathrm{20}}}{\mathrm{2}}{e}^{{i}\theta} \:{tel}\:{que}\:\begin{cases}{{cos}\theta=−\frac{\mathrm{3}}{\:\sqrt{\mathrm{20}}}}\\{{sin}\theta=\sqrt{\frac{\mathrm{11}}{\mathrm{20}}}}\end{cases} \\ $$$$\beta=\frac{\sqrt{\mathrm{20}}}{\mathrm{2}}{e}^{−{i}\theta} \: \\ $$$${U}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \left[\left(\frac{\sqrt{\mathrm{20}}}{\mathrm{2}}\right)^{{k}} {e}^{{ik}\theta} +\left(\frac{\sqrt{\mathrm{20}}}{\mathrm{2}}\right)^{{k}} {e}^{−{ik}\theta} \right] \\ $$$${U}_{{n}} =\frac{\mathrm{1}−\left(\frac{\sqrt{\mathrm{20}}}{\mathrm{2}}{e}^{{i}\theta} \right)^{{n}+\mathrm{1}} }{\mathrm{1}−\left(\frac{\sqrt{\mathrm{20}}}{\mathrm{2}}{e}^{{i}\theta} \right)}+\frac{\mathrm{1}−\left(\frac{\sqrt{\mathrm{20}}}{\mathrm{2}}{e}^{−{i}\theta} \right)^{{n}+\mathrm{1}} }{\mathrm{1}−\left(\frac{\sqrt{\mathrm{20}}}{\mathrm{2}}{e}^{−{i}\theta} \right)} \\ $$$${U}_{{n}} =\frac{\mathrm{1}−\frac{\sqrt{\mathrm{20}}}{\mathrm{2}}{e}^{−{i}\theta} −\left(\frac{\sqrt{\mathrm{20}}}{\mathrm{2}}{e}^{{i}\theta} \right)^{{n}+\mathrm{1}} +\left(\frac{\sqrt{\mathrm{20}}}{\mathrm{2}}\right)^{{n}+\mathrm{2}} {e}^{{i}\theta{n}} }{\mathrm{1}−\left(\frac{\sqrt{\mathrm{20}}}{\mathrm{2}}{e}^{−{i}\theta} \right)−\left(\frac{\sqrt{\mathrm{20}}}{\mathrm{2}}{e}^{{i}\theta} \right)+\frac{\mathrm{20}}{\mathrm{4}}} \\ $$$${to}\:{be}\:{continued}.... \\ $$