JEE Main & Advanced Mathematics Other Series Question Bank Critical Thinking

  • question_answer 15) The sum of the series \[1+\frac{3}{2\,!}+\frac{7}{3\,!}+\frac{15}{4\,!}+.....\text{to}\,\infty \] is  [Kerala (Engg.) 2005]

    A) \[e(e+1)\]

    B) \[e\,(1-e)\]

    C) \[3e-1\]

    D) \[3e\]

    E) (e) \[e\,(e-1)\]

    Correct Answer: E

    Solution :

    (e) \[1+\frac{3}{2\,!}+\frac{7}{3\,!}+\frac{15}{4\,!}+....\] \[=\,(1-1)+\left( \frac{2}{1!}-\frac{1}{1!} \right)+\left( \frac{{{2}^{2}}}{2\,!}-\frac{{{1}^{2}}}{2\,!} \right)+\left( \frac{{{2}^{3}}}{3\,!}-\frac{{{1}^{3}}}{3\,!} \right)+...\] \[=\left( 1+\frac{2}{1!}+\frac{{{2}^{2}}}{2!}+\frac{{{2}^{3}}}{3!}+.... \right)\]\[-\left( 1+\frac{1}{1!}+\frac{{{1}^{2}}}{2!}+\frac{{{1}^{3}}}{3!}+... \right)\] \[={{e}^{2}}-e\] = \[e(e-1)\].

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