Question

If $ 1+{{x}^{2}}=\sqrt{3}x, $ then $ \sum\limits_{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 i\,sin\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}}}=\left( \cos \frac{n\pi }{6}\pm i\sin \frac{n\pi }{6} \right. $ $ \left. -\cos \frac{n\pi }{6}\pm i\sin \frac{n\pi }{6} \right) $
$=\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\limits_{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 $