A) \[1\]
B) \[\omega \]
C) \[{{\omega }^{2}}\]
D) \[0\]
Correct Answer: D
Solution :
Given series \[1+\omega +{{\omega }^{2}}+.....+{{\omega }^{101}}\] is a GM series \[\therefore \] \[S=\frac{1({{\omega }^{102}}-1)}{\omega -1}\] \[=\frac{1({{({{\omega }^{2}})}^{34}}-1)}{\omega -1}=\frac{{{(1)}^{34}}-1}{\omega -1}\,\,\,\,\,(\because \,\,{{\omega }^{3}}=1)\] \[=0\]
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