Let α=-1+i32. If a=(1+α)∑k=0100α2k and b=∑k=0100α3k, then a and b are the roots of the quadratic equation:

α=ω, b=1+ω3+ω6+……+ω300=101×1=101a=(1+ω)1+ω2+ω4+……ω198+ω200=(1+ω)1-ω21011-ω2=(1+ω)(1-ω)1-ω2=1 Equation: x2-(101+1)x+(101)×1=0 ⇒x2-102x+101=0

Let α=-1+i32. If a=(1+α)∑k=0100α2k and b=∑k=0100α3k, then a and b are the roots of the quadratic equation:

α=ω, b=1+ω3+ω6+……+ω300=101×1=101a=(1+ω)1+ω2+ω4+……ω198+ω200=(1+ω)1-ω21011-ω2=(1+ω)(1-ω)1-ω2=1 Equation: x2-(101+1)x+(101)×1=0 ⇒x2-102x+101=0

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$\begin{array}{l} Let α = \frac{- 1 + i \sqrt{3}}{2} . If a = (1 + α) \sum_{k = 0}^{100} α^{2 k} and b = \sum_{k = 0}^{100} α^{3 k}, then a and b are the roots of the \\ quadratic equation: \end{array}$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$x^{2} - 101 x + 100 = 0$

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

$x^{2} + 102 x + 101 = 0$

$x^{2} + 101 x + 100 = 0$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} α = ω, b = 1 + ω^{3} + ω^{6} + \dots \dots + ω^{300} = 101 \times 1 = 101 \\ a = (1 + ω) (1 + ω^{2} + ω^{4} + \dots \dots ω^{198} + ω^{200}) \\ = (1 + ω) \frac{(1 - {(ω^{2})}^{101})}{1 - ω^{2}} = \frac{(1 + ω) (1 - ω)}{1 - ω^{2}} = 1 \\ Equation: x^{2} - (101 + 1) x + (101) \times 1 = 0 \Rightarrow x^{2} - 102 x + 101 = 0 \end{array}$