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 sum \[i+{{i}^{2}}+{{i}^{3}}+...\]to 1000 terms is equal to [Kerala (Engg.) 2002]

    A) 1

    B) - 1

    C) i

    D) 0

    Correct Answer: D

    Solution :

    \[i+{{i}^{2}}+{{i}^{3}}+.......\]up to 1000 terms \[=\frac{i(1-{{i}^{1000}})}{1-i}=\]\[\frac{i(1-{{({{i}^{4}})}^{250}})}{1-i}\]\[=\frac{i(1-1)}{1-i}=0\].

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