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(B) Given the quadratic equation $x^2 - 2x + 4 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{2 \pm \sqrt

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$2^{n+1} \cos \left( \frac{n\pi}{3} \right)$

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(B) Given the quadratic equation $x^2 - 2x + 4 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{2 \pm \sqrt{4 - 16}}{2} = \frac{2 \pm \sqrt{-12}}{2} = 1 \pm i\sqrt{3}$.We can write the roots in polar form: $\alpha = 1 + i\sqrt{3} = 2 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)$ and $\beta = 1 - i\sqrt{3} = 2 \left( \cos \frac{\pi}{3} - i \sin \frac{\pi}{3} \right)$.Using De Moivre's Theorem,$\alpha^n = 2^n \left( \cos \frac{n\pi}{3} + i \sin \frac{n\pi}{3} \right)$ and $\beta^n = 2^n \left( \cos \frac{n\pi}{3} - i \sin \frac{n\pi}{3} \right)$.Adding these,$\alpha^n + \beta^n = 2^n \left( \cos \frac{n\pi}{3} + i \sin \frac{n\pi}{3} + \cos \frac{n\pi}{3} - i \sin \frac{n\pi}{3} \right)$.$\alpha^n + \beta^n = 2^n \left( 2 \cos \frac{n\pi}{3} \right) = 2^{n+1} \cos \frac{n\pi}{3}$.

If $\alpha, \beta$ are the roots of the equation $x^2 - 2x + 4 = 0$,then the value of $\alpha^n + \beta^n$ is

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