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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 :
Option: 1 x^{2}+101x+100=0
Option: 2 x^{2}+102x+101=0
Option: 3 x^{2}-102x+101=0
Option: 4 x^{2}-101x+100=0
 

Answers (1)

best_answer

Cube roots of unity

\\\mathrm{\mathbf{\omega=\frac{-1+ i\sqrt{3}}{2}},\;\;\omega^2=\mathbf{\frac{-1- i\sqrt{3}}{2}}}

 

Sum of n-term of a GP

Let Sn be the sum of n terms of the G.P. with the first term ‘a’ and common ratio ‘r’. Then 

S_n=a( \frac{r^n-1}{r-1})

 

Now,

\alpha = \omega, 

b = 1 + \omega3 + \omega6 + ……\omega300= 101

a= (1+\omega) (1+\omega2+\omega4+\omega6.....+\omega200)

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

 

Now, equation with roots 1 and 101 is

x2 – (1+101)x + 101*1 = 0

x2 – 102x + 101 = 0

Correct Option (3)

Posted by

vishal kumar

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