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

  • question_answer
    The sum of the series \[\frac{{{1}^{2}}}{1\cdot 2\,!}+\frac{{{1}^{2}}+{{2}^{2}}}{2\cdot 3\,!}+\frac{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}{3\cdot 4\,!}+..+\frac{{{1}^{2}}+{{2}^{2}}+...+{{n}^{2}}}{n\cdot (n+1)\,!}+...\infty \]equals [AMU 2002]

    A) \[{{e}^{2}}\]

    B) \[\frac{1}{2}{{(e+{{e}^{-1}})}^{2}}\]

    C) \[\frac{3e-1}{6}\]

    D) \[\frac{4e+1}{6}\]

    Correct Answer: C

    Solution :

    \[{{T}_{n}}=\frac{n(n+1)(2n+1)}{6n(n+1)!}\] \[\therefore S=\frac{1}{6}\sum{\left[ \frac{2n+1}{(n)!} \right]}\] = \[\frac{1}{6}\sum{\left[ 2.\frac{1}{(n-1)!}+\frac{1}{(n)!} \right]}\]       = \[\frac{1}{6}[2.e+e-1]=\frac{1}{6}[3e-1]\].


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