A) \[50\]
B) \[-50\]
C) \[0\]
D) \[100\]
Correct Answer: C
Solution :
\[\therefore \]\[\sum\limits_{n=1}^{200}{{{i}^{n}}}=i+{{i}^{2}}+{{i}^{3}}+...+{{i}^{200}}\] \[(\because \]it is a \[GP\] series) \[=\frac{i(1-{{i}^{200}})}{1-i}=\frac{i(1-1)}{1-i}\] \[=0\]You need to login to perform this action.
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