Question

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

• A. $48$
• B. $-48$
• C. $\pm 48(\omega -{\omega }^2)$
• D. $1\pm 48\omega$

Answer 1

The correct option is B $-48$

Given, $1+x^2=\sqrt{3}x\Rightarrow x^2-\sqrt{3}x+1=0$

$\Rightarrow x=\frac{\sqrt{3}\pm \sqrt{3-4}}{2}=\frac{\sqrt{3}}{2}\pm \frac{i}{2}=\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}$

Now ${(x^n-\frac{1}{x^n})}^2=(x^{2n}+\frac{1}{x^{2n}}-2)=-2+2\cos \frac{n\pi }{3}$

$\Rightarrow \sum _{n=1}^{24}{(x^n-\frac{1}{x^n})}^2=\sum _{n=1}^{24}(-2+2\cos \frac{n\pi }{3})$

$=-48+2[\cos \frac{\pi }{3}+\cos \frac{2\pi }{3}+...+\cos \frac{24\pi }{3}]$

$=\frac{2[\cos (\frac{2\pi }{3}+\frac{23\pi }{6}).\sin \frac{24\pi }{6}]}{\sin \frac{\pi }{6}}-48=0-48=-48$