A) \[50\]
B) - 50
C) 0
D) 100
Correct Answer: C
Solution :
\[\sum\limits_{n=1}^{200}{{{i}^{n}}=i+{{i}^{2}}+{{i}^{3}}+....+{{i}^{200}}=\frac{i(1-{{i}^{200}})}{1-i}}\] (since G.P.) \[=\frac{i(1-1)}{1-i}=0\].
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