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(B) The given quadratic equation is $x^2-2x+4=0$. Using the quadratic formula,the roots are $\alpha, \beta = \frac{2 \pm \sqrt{4-16}}{2} = \frac{2 \pm

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$2^{n+1}$

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(B) The given quadratic equation is $x^2-2x+4=0$. Using the quadratic formula,the roots are $\alpha, \beta = \frac{2 \pm \sqrt{4-16}}{2} = \frac{2 \pm 2\sqrt{3}i}{2} = 1 \pm i\sqrt{3}$. Expressing the roots in polar form: $\alpha = 2\left(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}ight)$ and $\beta = 2\left(\cos \left(-\frac{\pi}{3}ight) + i \sin \left(-\frac{\pi}{3}ight)ight)$. Using De Moivre's Theorem,$\alpha^n + \beta^n = 2^n \left(\cos \frac{n\pi}{3} + i \sin \frac{n\pi}{3}ight) + 2^n \left(\cos \left(-\frac{n\pi}{3}ight) + i \sin \left(-\frac{n\pi}{3}ight)ight)$. Since $\cos(-	heta) = \cos 	heta$ and $\sin(-	heta) = -\sin 	heta$,we get: $\alpha^n + \beta^n = 2^n \left(2 \cos \frac{n\pi}{3}ight) = 2^{n+1} \cos \left(\frac{n\pi}{3}ight)$.

If $\alpha, \beta$ are the roots of the equation $x^2-2x+4=0$,then $\alpha^n+\beta^n = \ldots \cos \left(\frac{n\pi}{3}\right)$ for any $n \in N$.

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