Question

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

• A. 10
• B. 8
• C. 6
• D. 4

Answer 1

Answer: (B)

We have, $x^2-x+1=0$
$x=\frac{1\pm \sqrt{1-4}}{2}=\frac{1}{2}\pm \frac{\sqrt{3}i}{2}$
i.e., $x=\omega ,{\omega }^2$
$\therefore \sum _{n=1}^5{\left(x^n+\frac{1}{x^n}\right)}^2=\sum _{n=1}^5{\left({\omega }^n+\frac{1}{{\omega }^n}\right)}^2$
$=\sum _{n=1}^5\left[{\omega }^{2n}+\frac{1}{{\omega }^{2n}}+2\right]$
$=5\left(2\right)+{\omega }^2+\frac{1}{{\omega }^2}+{\omega }^4+\frac{1}{{\omega }^4}$
$+{\omega }^6+\frac{1}{{\omega }^6}+{\omega }^8+\frac{1}{{\omega }^8}+{\omega }^{10}+\frac{1}{{\omega }^{10}}$
$=10+\frac{{\omega }^4+1}{{\omega }^2}+\frac{{\omega }^8+1}{{\omega }^4}+\frac{{\omega }^{12}+1}{{\omega }^6}$
$+\frac{{\omega }^{16}+1}{{\omega }^8}+\frac{{\omega }^{20}+1}{{\omega }^{10}}$
$=10+\frac{\omega +1}{{\omega }^2}+\frac{{\omega }^2+1}{\omega }+\frac{1+1}{1}+\frac{\omega +1}{{\omega }^2}+\frac{{\omega }^2+1}{\omega }$
$=10-\frac{{\omega }^2}{{\omega }^2}=\frac{\omega }{\omega }+2-\frac{{\omega }^2}{{\omega }^2}-\frac{\omega }{\omega }$
$=10-1-1+2-1-1=8$.