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

  • question_answer
    \[{{\left[ 1+\frac{1}{2\,!}+\frac{1}{4\,!}+....\infty  \right]}^{2}}-{{\left[ 1+\frac{1}{3\,!}+\frac{1}{5\,!}+.....\infty  \right]}^{2}}=\]

    A)  0

    B) 1

    C) \[-1\]

    D) 2

    Correct Answer: B

    Solution :

    \[{{\left( \frac{e+{{e}^{-1}}}{2} \right)}^{2}}-{{\left( \frac{e-{{e}^{-1}}}{2} \right)}^{2}}=\frac{1}{4}\{4\,.\,e\,.\,{{e}^{-1}}\}=1\].


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