Let α = - 1 + i 3 2 . If a = (1 + α) ∑ _ k = 0 ^ 100 α ^ 2k and b =...

Given,

$\alpha = \dfrac{- 1 + i\sqrt{3}}{2}$

$a = \left(1 + \alpha\right)\mathop{∑}_{k = 0}^{100}{\alpha}^{2k}$ and

$b = \mathop{∑}_{k = 0}^{100}{\alpha}^{3k}$

Now

$\alpha = \omega; \omega = \dfrac{- 1 + i\sqrt{3}}{2}$

Using this $a$ and $b$ can be written as
$a = \left(1 + \omega\right)\left(1 + {\omega}^{2} + {\omega}^{4} + \ldots {\omega}^{198} + {\omega}^{200}\right)$
$= \left(1 + \omega\right)\dfrac{\left(1 - {\left({\omega}^{2}\right)}^{101}\right)}{1 - {\omega}^{2}} = \dfrac{\left(1 + \omega\right)\left(1 - \omega\right)}{1 - {\omega}^{2}} = 1$

Similarly,

$b = 1 + {\omega}^{3} + {\omega}^{6} + \ldots + {\omega}^{300} = 101$
So, the required quadratic equation is

${x}^{2} - \left(a + b\right)x + ab = 0$

$⇒ {x}^{2} - 102x + 101 = 0$.