A) 0
B) 48
C) \[-24\]
D) 24
E) \[-48\]
Correct Answer: E
Solution :
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, \[{{\sum\limits_{n=1}^{24}{\left( {{x}^{n}}-\frac{1}{{{x}^{n}}} \right)}}^{2}}\] \[=-4\left[ {{\sin }^{2}}\frac{\pi }{6}+{{\sin }^{2}}\frac{2\pi }{6}+....+{{\sin }^{2}}\frac{24\pi }{6} \right]\] \[=-4(12)=-48\]
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