Question

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

• A. $x^2 + 101x+100=0$
• B. $x^2 + 102x + 101=0$
• C. $x^2 - 102x + 101=0$
• D. $x^2 - 101x + 100=0$

Answer 1

Answer: (C)

$\alpha = \omega$
$a = \left(1 + \omega \right)\left(1 + \omega^{ 2} + \omega ^{4} + ..... + \omega ^{200}\right)$
$a = \left(1+\omega \right) \frac{\left(1-\left(\omega ^{2}\right)^{101}\right)}{1-\omega ^{2}} = 1$
$b = 1+\omega ^{3}+\omega ^{6}+\ldots\ldots+\omega ^{300} = 101$
$x^{2} - 102x + 101 = 0$