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(A) Given $\alpha$ is a root of $x^2+x+1=0$,so $\alpha = \omega$ or $\alpha = \omega^2$,where $\omega$ is a complex cube root of unity.Since $\frac{1}

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$11$

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(A) Given $\alpha$ is a root of $x^2+x+1=0$,so $\alpha = \omega$ or $\alpha = \omega^2$,where $\omega$ is a complex cube root of unity.Since $\frac{1}{\omega} = \omega^2$ and $\frac{1}{\omega^2} = \omega$,the expression $\left(\alpha^k+\frac{1}{\alpha^k}\right)^2$ simplifies to $\left(\omega^k+\omega^{2k}\right)^2 = \omega^{2k} + \omega^{4k} + 2\omega^{3k} = \omega^{2k} + \omega^k + 2$.We need to find $n$ such that $\sum_{k=1}^n (\omega^{2k} + \omega^k + 2) = 20$.This is $\sum_{k=1}^n \omega^{2k} + \sum_{k=1}^n \omega^k + 2n = 20$.If $n$ is a multiple of $3$,say $n=3m$,then $\sum_{k=1}^n \omega^{2k} = 0$ and $\sum_{k=1}^n \omega^k = 0$,so $2n = 20 \Rightarrow n = 10$ (not a multiple of $3$).If $n = 3m+1$,then $\sum_{k=1}^n \omega^{2k} = \omega^2$ and $\sum_{k=1}^n \omega^k = \omega$,so $\omega^2 + \omega + 2n = 20$ $\Rightarrow -1 + 2n = 20$ $\Rightarrow 2n = 21$ (no integer solution).If $n = 3m+2$,then $\sum_{k=1}^n \omega^{2k} = \omega^2 + \omega^4 = \omega^2 + \omega = -1$ and $\sum_{k=1}^n \omega^k = \omega + \omega^2 = -1$,so $-1 - 1 + 2n = 20$ $\Rightarrow 2n = 22$ $\Rightarrow n = 11$.Since $11 = 3(3) + 2$,this satisfies the condition.

If $\alpha$ is a root of the equation $x^2+x+1=0$ and $\sum_{k=1}^n\left(\alpha^k+\frac{1}{\alpha^k}\right)^2=20$,then $n$ is equal to

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