A) \[1\]
B) \[-1\]
C) \[i\]
D) \[0\]
Correct Answer: D
Solution :
Since, \[{{i}^{2}}=-1\] Let \[S=i+{{i}^{2}}+{{i}^{3}}+...+{{i}^{1000}}\] \[=\frac{i(1-{{i}^{1000}})}{1-i}\] (GP series) \[=\frac{i(1-1)}{1-i}=0\]You need to login to perform this action.
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