Question Number 227016 by Spillover last updated on 25/Dec/25
$${If}\:\alpha\:{and}\:.\beta\:\:{are}\:{root}\:{of}\:{the}\:{equation} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{4}=\mathrm{0}\:{and}\:{for}\:{n}\in\mathbb{N} \\ $$$${Show}\:{that}\: \\ $$$$\alpha^{{n}} +\beta^{{n}} =\mathrm{2}^{{n}+\mathrm{1}} \mathrm{cos}\:\left(\frac{{n}\pi}{\mathrm{3}}\right) \\ $$
Answered by Frix last updated on 25/Dec/25
![x^2 −2x+4=0 (x−1)^2 +3=0 x=1±i(√3)=2e^(±i(π/3)) α^n +β^n =2^n (e^(i((nπ)/3)) +e^(−i((nπ)/3)) )= [We know ((e^(iθ) +e^(−iθ) )/2)=cos θ] =2^(n+1) cos ((nπ)/3)](https://www.tinkutara.com/question/Q227017.png)
$${x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{4}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}=\mathrm{0} \\ $$$${x}=\mathrm{1}\pm\mathrm{i}\sqrt{\mathrm{3}}=\mathrm{2e}^{\pm\mathrm{i}\frac{\pi}{\mathrm{3}}} \\ $$$$\alpha^{{n}} +\beta^{{n}} =\mathrm{2}^{{n}} \left(\mathrm{e}^{\mathrm{i}\frac{{n}\pi}{\mathrm{3}}} +\mathrm{e}^{−\mathrm{i}\frac{{n}\pi}{\mathrm{3}}} \right)= \\ $$$$\:\:\:\:\:\left[\mathrm{We}\:\mathrm{know}\:\frac{\mathrm{e}^{\mathrm{i}\theta} +\mathrm{e}^{−\mathrm{i}\theta} }{\mathrm{2}}=\mathrm{cos}\:\theta\right] \\ $$$$=\mathrm{2}^{{n}+\mathrm{1}} \mathrm{cos}\:\frac{{n}\pi}{\mathrm{3}} \\ $$