let α and β roots of z^2 +3z+5=0 simlify U_n = Σ_(k=0) ^n (α^k +β^k ) and V_n =Σ


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Question Number 146902 by mathmax by abdo last updated on 16/Jul/21

$let α and β roots of z^{2} + 3 z + 5 = 0$ $simlify U_{n} = \sum_{k = 0}^{n} (α^{k} + β^{k})$ $and V_{n} = \sum_{k = 0}^{n} (\frac{1}{α^{k}} + \frac{1}{β^{k}})$
	Answered by ArielVyny last updated on 18/Jul/21
	$z^2 +3z+5=0→(z+(3/2))^2 −(9/4)+((20)/4)=0 (z+(3/2))^2 +((11)/4)=0 (z+(3/2)−i((√(11))/2))(z+(3/2)+i((√(11))/2))=0 then α=−(3/2)+i((√(11))/2) and β=−(3/2)−i((√(11))/2) U_n =Σ_(k=0) ^n (α^k +β^k ) ∣α∣=∣−(3/2)+i((√(11))/2)∣=(√((−(3/2))^2 +(((√(11))/2))^2 )) ∣α∣=(√((9/4)+((11)/4)))=((√(20))/2) α=((√(20))/2)(−((3/2)/((√(20))/2))+i(((√(11))/2)/((√(20))/2)))=((√(20))/2)(−(3/( (√(20))))+i((√(11))/( (√(20))))) α=((√(20))/2)e^(iθ) tel que { ((cosθ=−(3/( (√(20)))))),((sinθ=(√((11)/(20))))) :} β=((√(20))/2)e^(−iθ) U_n =Σ_(k=0) ^n [(((√(20))/2))^k e^(ikθ) +(((√(20))/2))^k e^(−ikθ) ] U_n =((1−(((√(20))/2)e^(iθ) )^(n+1) )/(1−(((√(20))/2)e^(iθ) )))+((1−(((√(20))/2)e^(−iθ) )^(n+1) )/(1−(((√(20))/2)e^(−iθ) ))) U_n =((1−((√(20))/2)e^(−iθ) −(((√(20))/2)e^(iθ) )^(n+1) +(((√(20))/2))^(n+2) e^(iθn) )/(1−(((√(20))/2)e^(−iθ) )−(((√(20))/2)e^(iθ) )+((20)/4))) to be continued....$
	$z^{2} + 3 z + 5 = 0 \to {(z + \frac{3}{2})}^{2} - \frac{9}{4} + \frac{20}{4} = 0$ ${(z + \frac{3}{2})}^{2} + \frac{11}{4} = 0$ $(z + \frac{3}{2} - i \frac{\sqrt{11}}{2}) (z + \frac{3}{2} + i \frac{\sqrt{11}}{2}) = 0$ $t h e n α = - \frac{3}{2} + i \frac{\sqrt{11}}{2} a n d β = - \frac{3}{2} - i \frac{\sqrt{11}}{2}$ $U_{n} = \sum_{k = 0}^{n} (α^{k} + β^{k})$ $∣ α ∣=∣ - \frac{3}{2} + i \frac{\sqrt{11}}{2} ∣= \sqrt{{(- \frac{3}{2})}^{2} + {(\frac{\sqrt{11}}{2})}^{2}}$ $∣ α ∣= \sqrt{\frac{9}{4} + \frac{11}{4}} = \frac{\sqrt{20}}{2}$ $α = \frac{\sqrt{20}}{2} (- \frac{\frac{3}{2}}{\frac{\sqrt{20}}{2}} + i \frac{\frac{\sqrt{11}}{2}}{\frac{\sqrt{20}}{2}}) = \frac{\sqrt{20}}{2} (- \frac{3}{\sqrt{20}} + i \frac{\sqrt{11}}{\sqrt{20}})$ $α = \frac{\sqrt{20}}{2} e^{i θ} t e l q u e {\begin{cases} c o s θ = - \frac{3}{\sqrt{20}} \\ s i n θ = \sqrt{\frac{11}{20}} \end{cases}$ $β = \frac{\sqrt{20}}{2} e^{- i θ}$ $U_{n} = \sum_{k = 0}^{n} [{(\frac{\sqrt{20}}{2})}^{k} e^{i k θ} + {(\frac{\sqrt{20}}{2})}^{k} e^{- i k θ}]$ $U_{n} = \frac{1 - {(\frac{\sqrt{20}}{2} e^{i θ})}^{n + 1}}{1 - (\frac{\sqrt{20}}{2} e^{i θ})} + \frac{1 - {(\frac{\sqrt{20}}{2} e^{- i θ})}^{n + 1}}{1 - (\frac{\sqrt{20}}{2} e^{- i θ})}$ $U_{n} = \frac{1 - \frac{\sqrt{20}}{2} e^{- i θ} - {(\frac{\sqrt{20}}{2} e^{i θ})}^{n + 1} + {(\frac{\sqrt{20}}{2})}^{n + 2} e^{i θ n}}{1 - (\frac{\sqrt{20}}{2} e^{- i θ}) - (\frac{\sqrt{20}}{2} e^{i θ}) + \frac{20}{4}}$ $t o b e c o n t i n u e d . . . .$
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