A) \[i\]
B) \[-i\]
C) \[1-i\]
D) \[1+i\]
E) 0
Correct Answer: E
Solution :
\[i-{{i}^{2}}+{{i}^{3}}-{{i}^{4}}+...-{{i}^{100}}\] This form a GP with common ratio\[(r=-1)\]. \[{{S}_{n}}=\frac{i\{1-{{(-i)}^{100}}\}}{1-(-i)}\] \[=\frac{i(1-{{i}^{100}})}{1+i}\] \[=\frac{i(1-1)}{1+i}\] \[(\because {{i}^{100}}=1)\] \[\therefore \] \[{{S}_{n}}=0\]You need to login to perform this action.
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