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. If $\alpha, \beta$ are the roots of the equation $x^2-2 x+4=0$, then $\alpha^n+\beta^n=\ldots \ldots$ $x \cos \left(\frac{n \pi}{3}\right)$ for any $n \in \mathbf{N}$
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Verified Answer
The correct answer is:
$2^{n+1}$
Given quadratic equation $x^2-2 x+4=0$ having roots $\alpha$ and $\beta$.
So, $\alpha, \beta=\frac{2 \pm \sqrt{4-16}}{2}=\frac{2 \pm 2 \sqrt{3} i}{2}=1 \pm \sqrt{3 i}$
$$
\begin{gathered}
=2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right) \text { and } \\
2\left(\cos \left(-\frac{\pi}{3}\right)+i \sin \left(-\frac{\pi}{3}\right)\right) \\
\therefore \alpha^{n+}+\beta^{n t}=2^{n t}\left[\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)^n\right. \\
\left.+\left(\cos \left(-\frac{\pi}{3}\right)+i \sin \left(-\frac{\pi}{3}\right)\right)^n\right] \\
=2^{n t}\left[2 \cos \frac{n \pi}{3}\right]=2^{n+1} \cos \left(\frac{n \pi}{3}\right)
\end{gathered}
$$
So, $\alpha, \beta=\frac{2 \pm \sqrt{4-16}}{2}=\frac{2 \pm 2 \sqrt{3} i}{2}=1 \pm \sqrt{3 i}$
$$
\begin{gathered}
=2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right) \text { and } \\
2\left(\cos \left(-\frac{\pi}{3}\right)+i \sin \left(-\frac{\pi}{3}\right)\right) \\
\therefore \alpha^{n+}+\beta^{n t}=2^{n t}\left[\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)^n\right. \\
\left.+\left(\cos \left(-\frac{\pi}{3}\right)+i \sin \left(-\frac{\pi}{3}\right)\right)^n\right] \\
=2^{n t}\left[2 \cos \frac{n \pi}{3}\right]=2^{n+1} \cos \left(\frac{n \pi}{3}\right)
\end{gathered}
$$
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