JEE Main 2020 — Mathematics Algebra

Given,

$$\alpha =\frac{-1+i\sqrt{3}}{2}$$α=2−1+i3​​

$\displaystyle a=(1+\alpha )\sum _{k=0}^{100}{\alpha }^{2k}$a=(1+α)k=0∑100​α2k and

$$b=\sum _{k=0}^{100}{\alpha }^{3k}$$b=k=0∑100​α3k

Now

$$\alpha =\omega ;\omega =\frac{-1+i\sqrt{3}}{2}$$α=ω;ω=2−1+i3​​

Using this $a$a and $b$b can be written as

$$a=(1+\omega )(1+{\omega }^{2}+{\omega }^{4}+\ldots {\omega }^{198}+{\omega }^{200})$$a=(1+ω)(1+ω2+ω4+…ω198+ω200)

$$=(1+\omega )\frac{(1-{({\omega }^{2})}^{101})}{1-{\omega }^{2}}=\frac{(1+\omega )(1-\omega )}{1-{\omega }^{2}}=1$$=(1+ω)1−ω2(1−(ω2)101)​=1−ω2(1+ω)(1−ω)​=1

Similarly,

$$b=1+{\omega }^{3}+{\omega }^{6}+\ldots +{\omega }^{300}=101$$b=1+ω3+ω6+…+ω300=101

So, the required quadratic equation is

$${x}^{2}-(a+b)x+ab=0$$x2−(a+b)x+ab=0

$\Rightarrow {x}^{2}-102x+101=0$⇒x2−102x+101=0.