A) \[{{2}^{n-1}}\cos \left( \frac{n\pi }{3} \right)\]
B) \[i{{2}^{n-1}}\sin \left( \frac{n\pi }{3} \right)\]
C) \[{{2}^{n+1}}\cos \left( \frac{n\pi }{3} \right)\]
D) \[i{{2}^{n+1}}\sin \left( \frac{n\pi }{3} \right)\]
Correct Answer: C
Solution :
Roots of the equation\[{{x}^{2}}-2x+4=0\]are\[\alpha \]and \[\beta \]. \[\therefore \]\[x=\frac{2\pm \sqrt{4-4\times 4}}{2}=1\pm \sqrt{-3}=1\pm i\sqrt{3}\] Let \[\alpha =1+i\sqrt{3},\beta =1-i\sqrt{3}\] \[\Rightarrow \] \[\alpha =2\left[ \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right]\] \[\beta =2\left[ \cos \frac{\pi }{3}-i\sin \frac{\pi }{3} \right]\] \[\therefore \]\[{{\alpha }^{n}}+{{\beta }^{n}}={{2}^{n}}{{\left[ \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right]}^{n}}\] \[+{{2}^{n}}{{\left[ \cos \frac{\pi }{3}-i\sin \frac{\pi }{3} \right]}^{n}}\] \[={{2}^{n}}\left[ \cos \frac{n\pi }{3}+i\sin \frac{n\pi }{3}+\cos \frac{n\pi }{3}-i\sin \frac{n\pi }{3} \right]\] \[={{2}^{n}}\left[ 2.\cos \frac{n\pi }{3} \right]={{2}^{n+1}}\cos \frac{n\pi }{3}\]
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