Question

If $1+x^2=\sqrt{3}x,$ then $\sum _{n=1}^{24}{\left(x^n-\frac{1}{x^n}\right)}^2$ is equal to

• A. 0
• B. 48
• C. $-24$
• D. 24
• E. $-48$

Answer 1

Answer: (E)

Given that, $1+x^2=\sqrt{3}x$
$\Rightarrow$ $x^2-\sqrt{3}x+1=0$
$\Rightarrow$ $x=\frac{\sqrt{3}\pm \sqrt{3-4}}{2}=\frac{\sqrt{3}\pm i}{2}$
$=\cos \frac{\pi }{6}\pm isin\frac{\pi }{6}$
$\Rightarrow$ $x^n=\cos \frac{n\pi }{6}\pm i\sin \frac{n\pi }{6}$ And $\frac{1}{x^n}=\cos \frac{n\pi }{6}\mp isin\frac{n\pi }{6}$
$\therefore$ $x^n-\frac{1}{x^n}=(\cos \frac{n\pi }{6}\pm i\sin \frac{n\pi }{6}$ $-\cos \frac{n\pi }{6}\pm i\sin \frac{n\pi }{6})$
$=\pm 2i\sin \frac{n\pi }{6}$
$\therefore$ ${\left(x^n-\frac{1}{x^n}\right)}^2=-4{\sin }^2\frac{n\pi }{6}$ Hence, ${\sum _{n=1}^{24}\left(x^n-\frac{1}{x^n}\right)}^2$
$=-4\left[{\sin }^2\frac{\pi }{6}+{\sin }^2\frac{2\pi }{6}+....+{\sin }^2\frac{24\pi }{6}\right]$
$=-4(12)=-48$