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