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

  • question_answer If \[{{i}^{2}}=-1\],  then the value of  \[\sum\limits_{n=1}^{200}{{{i}^{n}}}\]is [MP PET 1996]

    A) \[50\]

    B) - 50

    C) 0

    D) 100

    Correct Answer: C

    Solution :

      \[\sum\limits_{n=1}^{200}{{{i}^{n}}=i+{{i}^{2}}+{{i}^{3}}+....+{{i}^{200}}=\frac{i(1-{{i}^{200}})}{1-i}}\] (since G.P.)          \[=\frac{i(1-1)}{1-i}=0\].


You need to login to perform this action.
You will be redirected in 3 sec spinner