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If $\alpha$ is a root of $z^2-z+1=0$, then $\left(\alpha^{2014}+\frac{1}{\alpha^{2014}}\right)+\left(\alpha^{2015}+\frac{1}{\alpha^{2015}}\right)^2$
$$
\begin{aligned}
& +\left(\alpha^{2016}+\frac{1}{\alpha^{2016}}\right)^3+\left(\alpha^{2017}+\frac{1}{\alpha^{2017}}\right)^4+ \\
& \left(\alpha^{2018}+\frac{1}{\alpha^{2018}}\right)^5=
\end{aligned}
$$
Options:
$$
\begin{aligned}
& +\left(\alpha^{2016}+\frac{1}{\alpha^{2016}}\right)^3+\left(\alpha^{2017}+\frac{1}{\alpha^{2017}}\right)^4+ \\
& \left(\alpha^{2018}+\frac{1}{\alpha^{2018}}\right)^5=
\end{aligned}
$$
Solution:
2857 Upvotes
Verified Answer
The correct answer is:
8
$z^2-z+1=0$, then $z=-\omega$
$$
\begin{aligned}
& \text { So, }\left|(-\omega)^{2014}+\frac{1}{(-\omega)^{2014}}\right|+\left[(-\omega)^{2015}+\frac{1}{(-\omega)^{2015}}\right]^2 \\
& +\left[(-\omega)^{2016}+\frac{1}{(-\omega)^{2016}}\right]^3+\left[(-\omega)^{2017}+\frac{1}{(-\omega)^{2017}}\right]^4 \\
& +\left[(-\omega)^{2018}+\frac{1}{(-\omega)^{2018}}\right]^5 \\
& {\left[\begin{array}{r}
{\left[-\omega-\frac{1}{\omega}\right]+\left[\omega^2+\frac{1}{\omega^2}\right]^2+8+\left[-\omega+\frac{1}{-\omega}\right]^4+\left[\omega^2+\frac{1}{\omega^2}\right]^5} \\
=\left[+\omega+\omega^2\right]+\left[\omega^2+\omega\right]^2+8+\left[-\omega-\omega^2\right]^4+\left[\omega^2+\omega\right]^5 \\
=-1+1+8+1-1=8
\end{array}\right.}
\end{aligned}
$$
$$
\begin{aligned}
& \text { So, }\left|(-\omega)^{2014}+\frac{1}{(-\omega)^{2014}}\right|+\left[(-\omega)^{2015}+\frac{1}{(-\omega)^{2015}}\right]^2 \\
& +\left[(-\omega)^{2016}+\frac{1}{(-\omega)^{2016}}\right]^3+\left[(-\omega)^{2017}+\frac{1}{(-\omega)^{2017}}\right]^4 \\
& +\left[(-\omega)^{2018}+\frac{1}{(-\omega)^{2018}}\right]^5 \\
& {\left[\begin{array}{r}
{\left[-\omega-\frac{1}{\omega}\right]+\left[\omega^2+\frac{1}{\omega^2}\right]^2+8+\left[-\omega+\frac{1}{-\omega}\right]^4+\left[\omega^2+\frac{1}{\omega^2}\right]^5} \\
=\left[+\omega+\omega^2\right]+\left[\omega^2+\omega\right]^2+8+\left[-\omega-\omega^2\right]^4+\left[\omega^2+\omega\right]^5 \\
=-1+1+8+1-1=8
\end{array}\right.}
\end{aligned}
$$
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