A) \[{{2}^{2n}}\]
B) \[{{2}^{n}}\]
C) 1
D) 0
Correct Answer: A
Solution :
Given, \[{{\omega }^{3}}=1,1+\omega +{{\omega }^{2}}=0\] ...(i) Now, \[(1-\omega +{{\omega }^{2}}).(1-{{\omega }^{2}}+{{\omega }^{4}}).(1-{{\omega }^{4}}+{{\omega }^{8}}).\] \[(1-{{\omega }^{8}}+{{\omega }^{16}})........2n\] factors \[=(1-\omega +{{\omega }^{2}})\,\,(1-{{\omega }^{2}}+\omega )\] \[.(1-\omega +{{\omega }^{2}}).(1-{{\omega }^{2}}+\omega )......2n\] factors [from Eq. (i)] \[=(-\omega -\omega ).(-{{\omega }^{2}}-{{\omega }^{2}}).(-\omega -\omega ).(-{{\omega }^{2}}-{{\omega }^{2}})\] ... 2n factors [from Eq. (i)] \[=(-2\omega ).(-2{{\omega }^{2}}).(-2\omega ).(-2{{\omega }^{2}})....2n\]factors \[={{(-2)}^{2n}}{{(\omega )}^{n}}{{({{\omega }^{2}})}^{n}}\] \[={{(-1)}^{2n}}{{(2)}^{2n}}{{\omega }^{3n}}\] \[={{2}^{n}}.{{({{\omega }^{3}})}^{n}}={{2}^{2n}}{{(1)}^{n}}={{2}^{2n}}\] [from Eq. (i)]
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