A) \[50(1-i)\]
B) \[25i\]
C) \[25(1+i)\]
D) \[100(1-i)\]
Correct Answer: A
Solution :
Let \[S=i-2-3i+4+5i+.....+100{{i}^{100}}\] \[\Rightarrow \]\[S=i+2{{i}^{2}}+3{{i}^{3}}+4{{i}^{4}}+5{{i}^{5}}+........+100{{i}^{100}}\] Þ \[iS=\,\,\,\,\,\,\,\,\,{{i}^{2}}+2{{i}^{3}}+3{{i}^{4}}+4{{i}^{5}}+........+99{{i}^{100}}+100{{i}^{101}}\] \[\therefore \]\[S-iS=[i+{{i}^{2}}+{{i}^{3}}+{{i}^{4}}+......+{{i}^{100}}]-100{{i}^{101}}\] \[\Rightarrow \]\[S(1-i)=0-100{{i}^{101}}=-100i\] \[\therefore \]\[S=\frac{-100i}{1-i}=-50i(1+i)=-50(i-1)=50(1-i)\].
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