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

  • question_answer
    If \[S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+....+\infty ,\] then \[{{e}^{S}}=\]  [MP PET 1999]

    A) \[{{\log }_{e}}\left( \frac{4}{e} \right)\]

    B) \[\frac{4}{e}\]

    C) \[{{\log }_{e}}\left( \frac{e}{4} \right)\]

    D) \[\frac{e}{4}\]

    Correct Answer: B

    Solution :

    \[S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+.....\infty \]  \[=1-\frac{1}{2}-\frac{1}{2}+\frac{1}{3}+\frac{1}{3}-\frac{1}{4}-\frac{1}{4}+\frac{1}{5}+.....\infty \] \[=1+2\left[ -\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+.....\infty  \right]\] \[=1+2[{{\log }_{e}}2-1]=1+2{{\log }_{e}}2-2\] \[={{\log }_{e}}4-1={{\log }_{e}}4-{{\log }_{e}}e\] \[={{\log }_{e}}\left( \frac{4}{e} \right)\]; \[\therefore \,\,{{e}^{S}}=\frac{4}{e}\].


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