A) \[{{(\sqrt{2})}^{n-2}}\cos \left( \frac{n\pi }{4} \right)\]
B) \[{{(\sqrt{2})}^{n-2}}\sin \left( \frac{n\pi }{4} \right)\]
C) \[{{(\sqrt{2})}^{n+2}}\cos \left( \frac{n\pi }{4} \right)\]
D) \[{{(\sqrt{2})}^{n+2}}\sin \left( \frac{n\pi }{4} \right)\]
Correct Answer: C
Solution :
\[{{(1+i)}^{n}}+{{(1-i)}^{n}}\] \[={{(2)}^{n/2}}\left\{ \cos \frac{n\pi }{4}+i\sin \frac{n\pi }{4}+\cos \frac{n\pi }{4}-i\sin \frac{n\pi }{4} \right\}\] \[=\,{{2}^{\frac{n}{2}}}.\,2\cos \frac{n\pi }{4}={{2}^{\frac{n}{2}+1}}\cos \frac{n\pi }{4}={{(\sqrt{2})}^{n+2}}\cos \frac{n\pi }{4}\].
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