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JCECE Engineering JCECE Engineering Solved Paper-2002

  • question_answer
    If\[{{i}^{2}}=-1\]. Then sum\[i+{{i}^{2}}+{{i}^{3}}+....+{{i}^{1000}}\] term is equal to:

    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)                 \[=\frac{i(1-1)}{1-i}=0\]


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