Question

If $x^2-x+1=0$ then ${\sum }_{n=1}^5{(x^n+\frac{1}{x^n})}^2$=

• A. $8$
• B. $15$
• C. $5$
• D. $20$

Answer 1

The correct option is A $8$

Given equation $x^2-x+1=0\Rightarrow a=1b=-1c=1$

$\Rightarrow x=\frac{1\pm \sqrt{1-4}}{2}=\frac{1}{2}+\frac{\sqrt{3}}{2}i\Rightarrow x=\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}$

We know that $z^n=e^{in\theta }=\cos n\theta +i\sin n\theta$

Now, ${\sum }_{n=1}^5{(x^n+\frac{1}{x^n})}^2={\sum }_{n=1}^5{(e^{i\frac{n\pi }{3}}+\frac{1}{e^{i\frac{n\pi }{3}}})}^2$

$\Rightarrow {\sum }_{n=1}^5{(x^n+\frac{1}{x^n})}^2={\sum }_{n=1}^5[(\cos \frac{n\pi }{3}+i\sin \frac{n\pi }{3})+\cos \frac{n\pi }{3}-i\sin \frac{n\pi }{3}]={\sum }_{n=1}^5{\left(2\cos \frac{n\pi }{3}\right)}^2$

$=4{\sum }_{n=1}^5{\left(\cos \frac{n\pi }{3}\right)}^2=4[{\cos }^2\frac{\pi }{3}+{\cos }^2\frac{2\pi }{3}+{\cos }^2\frac{3\pi }{3}+{\cos }^2\frac{4\pi }{3}+{\cos }^2\frac{5\pi }{3}]=4[\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+1]=4[1+1]=4\left(2\right)=8$

$\therefore {\sum }_{n=1}^5{(x^n+\frac{1}{x^n})}^2=8$

$\therefore$ option A is correct.