Let α=−1+i32.if a=(1+α)∑k=0100 α2k&b=∑k=0100 α3k then a and b are the roots of the quadratic equation

Itisgiventhat α=−1+i32, then α2+α+1=0 and α3=1 So, a=(1+α)∑k=0100 α2k=−α2∑k=0100 α2k=−∑k=0100 α2(k+1) =−α2+α4+α6+α8+…+α202 =−α2α2101−1α2−1 (sum of GP) =−α2α202−1α2−1=−α2(α−1)α2−1, ∵α3=1 =−α2α+1=−α3α2+α=1 and, b=∑k=0100 α3k=∑k=0100 α3k=∑k=0100 1=101Now, equation of quadratic equation having roots to'a' and 'b' isx2−(a+b)x+ab=0⇒x2−102x+101=0Hence, option (c) is correct.

Let α=−1+i32.if a=(1+α)∑k=0100 α2k&b=∑k=0100 α3k then a and b are the roots of the quadratic equation

Itisgiventhat α=−1+i32, then α2+α+1=0 and α3=1 So, a=(1+α)∑k=0100 α2k=−α2∑k=0100 α2k=−∑k=0100 α2(k+1) =−α2+α4+α6+α8+…+α202 =−α2α2101−1α2−1 (sum of GP) =−α2α202−1α2−1=−α2(α−1)α2−1, ∵α3=1 =−α2α+1=−α3α2+α=1 and, b=∑k=0100 α3k=∑k=0100 α3k=∑k=0100 1=101Now, equation of quadratic equation having roots to'a' and 'b' isx2−(a+b)x+ab=0⇒x2−102x+101=0Hence, option (c) is correct.

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Let $α = \frac{- 1 + i \sqrt{3}}{2} .$ if $a = (1 + α) \sum_{k = 0}^{100} α^{2 k} & b = \sum_{k = 0}^{100} α^{3 k}$ then a and b are the roots of the quadratic equation

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

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

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

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

answer is C.

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Detailed Solution

$Itisgiventhat α = \frac{- 1 + i \sqrt{3}}{2}, then α^{2} + α + 1 = 0 and$ $α^{3} = 1$
$So, a = (1 + α) \sum_{k = 0}^{100} α^{2 k} = (- α^{2}) \sum_{k = 0}^{100} α^{2 k} = - \sum_{k = 0}^{100} α^{2 (k + 1)} = - [α^{2} + α^{4} + α^{6} + α^{8} + \dots + α^{202}] = - \frac{α^{2} ({(α^{2})}^{101} - 1)}{α^{2} - 1} (sum of GP) = - \frac{α^{2} (α^{202} - 1)}{α^{2} - 1} = - \frac{α^{2} (α - 1)}{α^{2} - 1}, (∵ α^{3} = 1) = - \frac{α^{2}}{α + 1} = - \frac{α^{3}}{α^{2} + α} = 1 and, b = \sum_{k = 0}^{100} α^{3 k} = \sum_{k = 0}^{100} {(α^{3})}^{k} = \sum_{k = 0}^{100} 1 = 101$
Now, equation of quadratic equation having roots to'a' and 'b' is
$x^{2} - (a + b) x + ab = 0 \Rightarrow x^{2} - 102 x + 101 = 0$
Hence, option (c) is correct.