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.

Let α = -1+i√3 2. If a=(1+α )Σ _k=0^100… - JEE Main 2020 Mathematics PYQ

JEE Main 2020 — Mathematics Algebra

2020mcqhard

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

Official previous-year question

Held on 8 Jan 2020 · Verified 6 Jul 2026.

Options

A
${x}^{2}+101x+100=0$
B
${x}^{2}-102x+101=0$
C
${x}^{2}-101x+100=0$
D
${x}^{2}+102x+101=0$

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Solution

Given,

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

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

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

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

Similarly,

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

So, the required quadratic equation is

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

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

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