JEE Mains · Maths · STD 11 - 4.1 complex nubers
If \(\alpha\) is a root of the equation \(x^2+x+1=0\) and \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\left(\alpha^{\mathrm{k}}+\frac{1}{\alpha^{\mathrm{k}}}\right)^2=20\), then n is equal to
- A 5
- B 7
- C 11
- D 15
Answer & Solution
Correct Answer
(C) 11
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \alpha=\omega \\ & \therefore\left(\omega^k+\frac{1}{\omega^k}\right)^2=\omega^{2 k}+\frac{1}{\omega^{2 k}}+2 \\ & =\omega^{2 k}+\omega^k+2 \quad \because \omega^{3 k}=1\end{aligned}\)…
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