A) \[2n\]
B) \[{{2}^{2n}}\]
C) \[1\]
D) \[-{{2}^{2n}}\]
Correct Answer: B
Solution :
\[(1-\omega +{{\omega }^{2}})\,(1-{{\omega }^{2}}+{{\omega }^{4}})\,(1-{{\omega }^{4}}+{{\omega }^{8}})\] \[(1-{{\omega }^{8}}+{{\omega }^{16}}).....\] upto \[2n\] \[=(1-\omega +{{\omega }^{2}})\,(1-{{\omega }^{2}}+{{\omega }^{2}}.\omega )\] \[(1-{{\omega }^{3}}.\omega +{{\omega }^{6}}.{{\omega }^{2}})\] \[(1-{{\omega }^{6}}.{{\omega }^{2}}+{{\omega }^{15}}.\omega ).....\] upto \[2n\] \[=(1-\omega +{{\omega }^{2}})(1-{{\omega }^{2}}+\omega )\] \[(1-\omega +{{\omega }^{2}})(1-{{\omega }^{2}}+\omega )....\] upto \[2n\] \[=[(-2\omega )\,(-2{{\omega }^{2}})\,(-2{{\omega }^{2}}).....\] uptp \[2n\] \[=[(-2\omega )\,(-2{{\omega }^{2}})]\times [(-2\omega )\,(-2{{\omega }^{2}})]\] \[\times ....\] upto \[2n\] \[=({{2}^{2}}\,{{\omega }^{3}})\times ({{2}^{2}}{{\omega }^{3}})\times ....\]upto n \[=[{{2}^{2}}\times {{2}^{2}}\times {{2}^{2}}\times ...upto\,n]={{2}^{2n}}\]
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