Question

If $x^2-x+1=0,$ then value of $\sum _{n=1}^5{\left(x^n+\frac{1}{x^n}\right)}^2$ is.

• A. 8
• B. 10
• C. 12
• D. none

Answer 1

The correct option is A 8

We have,

$x^2-x+1=0-\omega ,-{\omega }^2$

$x=\frac{1\pm \sqrt{3i}}{2}$

$\begin{array}{c}{\left(x+\frac{1}{x}\right)}^2+{\left(x^2+\frac{1}{x^2}\right)}^2+{\left(x^3+\frac{1}{x^3}\right)}^2+{\left(x^4+\frac{1}{x^4}\right)}^2+{\left(x^5+\frac{1}{x^5}\right)}^2\\ {\left(-\omega -\frac{1}{\omega }\right)}^2+{\left({\omega }^2+\frac{1}{{\omega }^2}\right)}^2+{\left(-{\omega }^3-\frac{1}{{\omega }^3}\right)}^2+{\left({\omega }^4+\frac{1}{{\omega }^4}\right)}^2+{\left(-{\omega }^5-\frac{1}{{\omega }^5}\right)}^2\\ =\frac{{\left(1+{\omega }^2\right)}^2}{{\omega }^2}+\frac{{\left({\omega }^4+1\right)}^2}{{\omega }^4}+\frac{{\left({\omega }^6+1\right)}^2}{{\omega }^6}+\frac{{\left({\omega }^8+1\right)}^2}{{\omega }^8}+\frac{{\left({\omega }^{10}+1\right)}^2}{{\omega }^{10}}\\ =1+1+4+1+1{\omega }^3=1,1+{\omega }^2=-\omega ,{\omega }^6=1\end{array}$

Then,

Option $A$ is correct answer.