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JEE Main & Advanced Mathematics Sequence & Series Question Bank Exponential series

  • question_answer
    In  the expansion of  \[\frac{{{e}^{4x}}-1}{{{e}^{2x}}}\], the coefficient of \[{{x}^{2}}\] is

    A) \[\frac{1}{2}\]

    B) 1

    C) 0

    D) None of these

    Correct Answer: C

    Solution :

    \[\frac{{{e}^{4x}}-1}{{{e}^{2x}}}={{e}^{2x}}-{{e}^{-2x}}=2\left\{ 2x+\frac{{{(2x)}^{3}}}{3\ !}+\frac{{{(2x)}^{5}}}{5\ !}+..... \right\}\] The coefficient of\[{{x}^{2}}=0\].


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