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JEE Mains · Maths · STD 11 - 4.1 complex nubers

यदि \(\alpha\) समीकरण \(x^2+x+1=0\) का एक मूल है और \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\left(\alpha^{\mathrm{k}}+\frac{1}{\alpha^{\mathrm{k}}}\right)^2=20\), तो n = __________

  1. A 5
  2. B 7
  3. C 11
  4. D 15
Verified Solution

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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