If (1+i)2n+1(1−i)2n−1,n∈N is equal to __________(where i=-1)

(1+i)2n+1(1−i)2n−1=(1+i1−i)2n(1+i)(1−i) =((1+i1−i)2)n(1−i2) =(−1)n(1+1)=2(−1)n

If (1+i)2n+1(1−i)2n−1,n∈N is equal to __________(where i=-1)

(1+i)2n+1(1−i)2n−1=(1+i1−i)2n(1+i)(1−i) =((1+i1−i)2)n(1−i2) =(−1)n(1+1)=2(−1)n

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If $\frac{{(1 + i)}^{2 n + 1}}{{(1 - i)}^{2 n - 1}}, n \in N$ is equal to __________(where $i = \sqrt{- 1}$ )

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$2 i$

$2 {(- 1)}^{n}$

$\sqrt{2} + i \sqrt{2}$

${(- 1)}^{n}$

answer is B.

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Detailed Solution

$\frac{{(1 + i)}^{2 n + 1}}{{(1 - i)}^{2 n - 1}} = {(\frac{1 + i}{1 - i})}^{2 n} (1 + i) (1 - i) = {({(\frac{1 + i}{1 - i})}^{2})}^{n} (1 - i^{2}) = {(- 1)}^{n} (1 + 1) = 2 {(- 1)}^{n}$

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If (1+i)2n+1(1−i)2n−1,n∈N is equal to __________(where i=-1)

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If $\frac{{(1 + i)}^{2 n + 1}}{{(1 - i)}^{2 n - 1}}, n \in N$ is equal to __________(where $i = \sqrt{- 1}$ )