Solve the following
Question:

Let $\alpha=\frac{-1+i \sqrt{3}}{2} .$ If $a=(1+\alpha) \sum_{k=0}^{100} \alpha^{2 k}$ and $b=\sum_{k=0}^{100} \alpha^{3 k}$,

then $a$ and $b$ are the roots of the quadratic equation:

  1. (1) $x^{2}+101 x+100=0$

  2. (2) $x^{2}-102 x+101=0$

  3. (3) $x^{2}-101 x+100=0$

  4. (4) $x^{2}+102 x+101=0$


Correct Option: , 2

Solution:

Let $\alpha=\omega, b=1+\omega^{3}+\omega^{6}+\ldots . .=101$

$a=(1+\omega)\left(1+\omega^{2}+\omega^{4}+\ldots . . \omega^{198}+\omega^{200}\right)$

$=(1+\omega) \frac{\left(1-\left(\omega^{2}\right)^{101}\right)}{1-\omega^{2}}=\frac{(\omega+1)\left(\omega^{202}-1\right)}{\left(\omega^{2}-1\right)}$

$\Rightarrow a=\frac{(1+\omega)(1-\omega)}{1-\omega^{2}}=1$

Required equation $=x^{2}-(101+1) x+(101) \times 1=0$

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

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