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

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Aicharya · Accepted Answer

$\begin{aligned} & \alpha=\omega \ & 	herefore\left(\omega^k+\frac{1}{\omega^k}ight)^2=\omega^{2 k}+\frac{1}{\omega^{2 k}}+2 \ & =\omega^{2 k}+\omega^k+2 \quad \because \omega^{3 k}=1\end{aligned}$ $\begin{aligned} & 	herefore \sum_{\mathrm{k}=1}^{\mathrm{n}}\left(\omega^{2 \mathrm{k}}+\omega^{\mathrm{k}}+2ight)=20 \ & \Rightarrow\left(\omega^2+\omega^4+\omega^6+\ldots+\omega^{2 \mathrm{n}}ight)+\left(\omega+\omega^2+\omega^3+\ldots+ight. \ & \left.\omega^{\mathrm{n}}ight)+2 \mathrm{n}=20 \end{aligned}$ Now if $n=3 m, \quad m \in I$ Then $0+0+2 \mathrm{n}=20 \Rightarrow \mathrm{n}=10$ (not satisfy) if $n=3 m+1$, then $\begin{aligned} & \omega^2+\omega+2 \mathrm{n}=20 \ & -1+2 \mathrm{n}=20 \Rightarrow \mathrm{n}=\frac{21}{2}(	ext { not possible }) \end{aligned}$ if $n=3 m+2$, $\begin{aligned} & \left(\omega^8+\omega^{10}ight)+\left(\omega^4+\omega^5ight)+2 \mathrm{n}=20 \ & \Rightarrow\left(\omega^2+\omegaight)+\left(\omega+\omega^2ight)+2 \mathrm{n}=20 \ & 2 \mathrm{n}=22 \ & \mathrm{n}=11 	ext { satisfy } \mathrm{n}=3 \mathrm{~m}+2 \ & 	herefore \mathrm{n}=11 \end{aligned}$

If $\alpha$ is a root of the equation $x^2+x+1=0$ and…

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