Question

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:

• 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

The correct option is C $x^2-102x+101=0$

Clearly, $\alpha =\frac{-1+i\sqrt{3}}{2}=\omega$
where ${\omega }^3=1$

$a=(1+\alpha )\sum _{k=0}^{100}{\alpha }^{2k}$
$\Rightarrow a=(1+\alpha )[1+{\alpha }^2+{\alpha }^4+...+{\alpha }^{200}]$
$\Rightarrow a=(1+\alpha )\left[\frac{1-({\alpha }^2{)}^{101}}{1-{\alpha }^2}\right]$

$\Rightarrow a=\left[\frac{1-({\omega }^2{)}^{101}}{1-\omega }\right]=\left[\frac{1-\omega }{1-\omega }\right]=1$

$b=\sum _{k=0}^{100}{\alpha }^{3k}=1+{\alpha }^3+{\alpha }^6+...+{\alpha }^{300}$

$\Rightarrow b=1+{\omega }^3+{\omega }^6+...+{\omega }^{300}$
$\Rightarrow b=101$

Required quadratic equation is $x^2-102x+101=0$