Let α=−1+i32, if a=(1+α)∑k=0100α2k and b=∑k=1100α6k Then the Quadratic equation where roots are 2a, 2b is

α=−1+i32=w b=w6+w12+........=100 a=(1+w)(1+w2+w4+......+w200) x2−(202)x+400=0

Let α=−1+i32, if a=(1+α)∑k=0100α2k and b=∑k=1100α6k Then the Quadratic equation where roots are 2a, 2b is

α=−1+i32=w b=w6+w12+........=100 a=(1+w)(1+w2+w4+......+w200) x2−(202)x+400=0

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Let $α = \frac{- 1 + i \sqrt{3}}{2},$ if $a = (1 + α) \sum_{k = 0}^{100} α^{2 k}$ and $b = \sum_{k = 1}^{100} α^{6 k}$ Then the Quadratic equation where roots are 2a, 2b is

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$x^{2} + 202 x + 400 = 0$

$x^{2} + 202 x - 400 = 0$

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

$x^{2} - 202 x + 400 = 0$

answer is D.

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$α = \frac{- 1 + i \sqrt{3}}{2} = w b = w^{6} + w^{12} + ........ = 100 a = (1 + w) (1 + w^{2} + w^{4} + ...... + w^{200}) x^{2} - (202) x + 400 = 0$

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Let α=−1+i32, if a=(1+α)∑k=0100α2k and b=∑k=1100α6k Then the Quadratic equation where roots are 2a, 2b is

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Let $α = \frac{- 1 + i \sqrt{3}}{2},$ if $a = (1 + α) \sum_{k = 0}^{100} α^{2 k}$ and $b = \sum_{k = 1}^{100} α^{6 k}$ Then the Quadratic equation where roots are 2a, 2b is