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JCECE Engineering JCECE Engineering Solved Paper-2003

  • question_answer
    If\[{{i}^{2}}=-1\], then the value of\[\sum\limits_{n=1}^{200}{{{i}^{n}}}\]is:

    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)                 \[=\frac{i(1-{{i}^{200}})}{1-i}=\frac{i(1-1)}{1-i}\]                 \[=0\]


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