JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(\alpha=\frac{-1+i \sqrt{3}}{2} .\) If \(a=(1+\alpha) \sum\limits_{k=0}^{100} \alpha^{2 k}\) and \(\mathrm{b}=\sum\limits_{\mathrm{k}=0}^{100} \alpha^{3 \mathrm{k}},\) then a and \(\mathrm{b}\) are the roots of the quadratic equation
- A \(\mathrm{x}^{2}-102 \mathrm{x}+101=0\)
- B \(\mathrm{x}^{2}+101 \mathrm{x}+100=0\)
- C \(x^{2}-101 x+100=0\)
- D \(x^{2}+102 x+101=0\)
Answer & Solution
Correct Answer
(A) \(\mathrm{x}^{2}-102 \mathrm{x}+101=0\)
Step-by-step Solution
Detailed explanation
\(\alpha=\omega\) \(a=(1+\omega)\left(1+\omega^{2}+\omega^{4}+\ldots \ldots+\omega^{200}\right)\) \(a=(1+\omega) \frac{\left(1-\left(\omega^{2}\right)^{101}\right)}{1-\omega^{2}}=1\) \(\mathrm{b}=1+\omega^{3}+\omega^{6}+\ldots \ldots+\omega^{300}=101\) \(x^{2}-102 x+101=0\)
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