Question

Let $\alpha =\frac{(-1+\sqrt{3}i)}{2}$ . 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:

• 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

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

Find the required quadratic equation:

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

$\Rightarrow$ $\alpha =\omega$

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

$$\begin{gathered} =(1+\alpha )[1+{\alpha }^2+{\alpha }^4+\cdots .+{\alpha }^{200}] \\ =(1+\alpha )\left[\frac{{({\alpha }^2)}^{101}-1}{({\alpha }^2-1)}\right] \\ =\frac{\left(\omega +1\right)\left({\omega }^{202}-1\right)}{({\omega }^2-1)} \\ =\frac{\left(\omega +1\right)\left(\omega -1\right)}{\left(\omega +1\right)\left(\omega -1\right)} \\ =1 \end{gathered}$$

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

$$\begin{gathered} =1+{\alpha }^3+{\alpha }^6+.....+{\alpha }^{300} \\ =1+{\omega }^3+{\omega }^6+.....+{\omega }^{300} \\ =1+1+\hspace{0.17em}1+.....+101times \\ =101 \end{gathered}$$

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

$\Rightarrow$ ${\mathit{x}}^{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{102}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\mathbf{101}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$

Hence, Option ‘C’ is Correct.