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

Given,

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

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

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

Now

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

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

Similarly,

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

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

$\Rightarrow x^2-102x+101=0$.