Let alpha = frac{-1 + i sqrt{3}}{2}. If a = ( | vedclass

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(A) Given $\alpha = \frac{-1 + i \sqrt{3}}{2} = \omega$,where $\omega$ is the cube root of unity,satisfying $\omega^{3} = 1$ and $1 + \omega + \omega^

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$x^{2} - 102x + 101 = 0$

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(A) Given $\alpha = \frac{-1 + i \sqrt{3}}{2} = \omega$,where $\omega$ is the cube root of unity,satisfying $\omega^{3} = 1$ and $1 + \omega + \omega^{2} = 0$.For $a = (1 + \alpha) \sum_{k=0}^{100} \alpha^{2k} = (1 + \omega) (1 + \omega^{2} + \omega^{4} + \dots + \omega^{200})$.Since $\omega^{3} = 1$,the sequence $1, \omega^{2}, \omega^{4}, \dots$ repeats every three terms as $1, \omega^{2}, \omega$. The sum of three consecutive terms is $1 + \omega^{2} + \omega = 0$.There are $101$ terms in the sum. The sum of the first $99$ terms is $0$. The remaining two terms are the $100^{th}$ and $101^{st}$ terms: $1 + \omega^{2}$.Thus,$a = (1 + \omega)(1 + \omega^{2}) = 1 + \omega^{2} + \omega + \omega^{3} = 0 + 1 = 1$.For $b = \sum_{k=0}^{100} \alpha^{3k} = \sum_{k=0}^{100} (\omega^{3})^{k} = \sum_{k=0}^{100} (1)^{k} = 101$.The quadratic equation with roots $a = 1$ and $b = 101$ is given by $(x - a)(x - b) = 0$,which is $(x - 1)(x - 101) = x^{2} - 102x + 101 = 0$.

Let $\alpha = \frac{-1 + i \sqrt{3}}{2}$. If $a = (1 + \alpha) \sum_{k=0}^{100} \alpha^{2k}$ and $b = \sum_{k=0}^{100} \alpha^{3k}$,then $a$ and $b$ are the roots of the quadratic equation:

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