Question

If $x^2-x\sqrt{3}+1=0$, then $\sum _{n=1}^{36}{(x^n-\frac{1}{x^n})}^{2}dx$ is equal to

• A. $0$
• B. $-72$
• C. $72$
• D. none of these

Answer 1

Answer: (B)

$-72$

Given $x^2-x\sqrt{3}+1=0$
$\therefore x=\frac{\sqrt{3}\pm i}{2}=\frac{\sqrt{3}}{2}\pm \frac{i}{2}$ $x=\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}$
$\Rightarrow x^{2n}=\cos \frac{n\pi }{3}+i\sin \frac{n\pi }{3}$
$\therefore {(x^n-\frac{1}{x^n})}^2=x^{2n}+\frac{1}{x^{2n}}-2=x^{2n}+x^{-2n}-2=-2+2\cos \frac{n\pi }{3}$
Now
$\sum _{n=1}^{36}(-2+2cos\frac{n\pi }{3})$ $=-2\times 36+2[\cos \left(\frac{\pi }{3}\right)+\cos \frac{2\pi }{3}+...+\cos \frac{36\pi }{3}]$
$=-72+2\frac{\cos (\frac{\pi }{3}+\frac{35\pi }{6}).\sin \left(\frac{36\pi }{3\times 2}\right)}{\sin \frac{\pi }{6}}$ $=-72+0$, $\because [\cos \alpha (\alpha +\beta )+...\text{upto n terms}=\frac{\sin \frac{n\beta }{2}}{\sin \frac{\beta }{2}}\cos (\alpha +\frac{(n-1)\beta }{2})]$
Hence, option B is correct.