A) \[{{2}^{n}}\]
B) \[{{2}^{2n}}\]
C) 0
D) 1
Correct Answer: B
Solution :
\[{{\omega }^{4}}=\omega ,{{\omega }^{8}}={{\omega }^{2}}\]etc. 3rd, 5th, 7th factors are each equal to 1st and 4th, 6th, 8th factors are each equal to 2nd. L.H.S \[=(-2\omega )(-2{{\omega }^{2}})(-2\omega )(-2{{\omega }^{2}}).....\]to \[2n\] factors \[=({{2}^{2}}{{\omega }^{3}})({{2}^{2}}{{\omega }^{3}})......\]to\[n\]factors \[={{({{2}^{2}})}^{n}}={{2}^{2n}}\]
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