Question

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

Answer 1

Answer: 8.00

$x^2-x+1=0$
$\Rightarrow x=\frac{1\pm \sqrt{3}i}{2}=-\omega ,-{\omega }^2$
$\therefore \sum _{n=1}^5\left(x^{2n}+\frac{1}{x^{2n}}+2\right)$
$\Rightarrow \left(x^2+\frac{1}{x^2}+2\right)+\left(x^4+\frac{1}{x^4}+2\right)+$
$\left(x^6+\frac{1}{x^6}+2\right)+\left(x^8+\frac{1}{x^8}+2\right)+$
$\left(x^{10}+\frac{1}{x^{10}}+2\right)$
$\Rightarrow \left({\omega }^2+{\omega }^4+{\omega }^6+{\omega }^8+{\omega }^{10}\right)+$
$\left(\frac{1}{{\omega }^2}+\frac{1}{{\omega }^4}+\frac{1}{{\omega }^6}+\frac{1}{{\omega }^8}+\frac{1}{{\omega }^{10}}\right)+10$
$\Rightarrow -1-1+10=8.$