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JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Integral power of iota, Algebraic operations and Equality of complex numbers

  • question_answer
    The value of the sum \[\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}\], where \[i=\sqrt{-1}\], equals [IIT 1998]

    A) \[i\]

    B) \[i-1\]

    C) \[-i\]

    D) 0

    Correct Answer: B

    Solution :

      \[\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}\]\[=(i+{{i}^{2}}+{{i}^{3}}+....+{{i}^{13}})+({{i}^{2}}+{{i}^{3}}+....+{{i}^{14}})\] \[=\frac{i(1-{{i}^{13}})}{1-i}+\frac{{{i}^{2}}(1-{{i}^{13}})}{1-i}\] \[=i\,\left( \frac{1-i}{1-i} \right)+\frac{{{i}^{2}}(1-i)}{(1-i)}\]\[=i+{{i}^{2}}=i-1\].


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