2n(1−i)2n+(1+i)2n2n,n∈I is equal to

E=2n(−2i)n+(2i)n2n ∵(1+i)2=2i =1(−1)nin+in1=1+(−1)ni2n(−1)nin=1+1(−1)nin=2(−i)n

2n(1−i)2n+(1+i)2n2n,n∈I is equal to

E=2n(−2i)n+(2i)n2n ∵(1+i)2=2i =1(−1)nin+in1=1+(−1)ni2n(−1)nin=1+1(−1)nin=2(−i)n

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$\frac{2^{n}}{(1 - i)^{2 n}} + \frac{(1 + i)^{2 n}}{2^{n}}, n \in I$ is equal to

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$[1 + (- 1)^{n}] i^{n}$

$\frac{2}{(- i)^{n}}$

answer is D.

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Detailed Solution

$E = \frac{2^{n}}{(- 2 i)^{n}} + \frac{(2 i)^{n}}{2^{n}} ∵ (1 + i)^{2} = 2 i \begin{array}{l} = \frac{1}{(- 1)^{n} i^{n}} + \frac{i^{n}}{1} = \frac{1 + (- 1)^{n} {(i^{2})}^{n}}{(- 1)^{n} i^{n}} \\ = \frac{1 + 1}{(- 1)^{n} i^{n}} = \frac{2}{(- i)^{n}} \end{array}$