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

  • question_answer
    \[\frac{4}{1\,.\,3}-\frac{6}{2.4}+\frac{12}{5\,.\,7}-\frac{14}{6\,.\,8}+.....\infty =\]

    A) \[{{\log }_{e}}3\]

    B) \[{{\log }_{e}}2\]

    C) \[2\,{{\log }_{e}}2\]

    D) None of these

    Correct Answer: B

    Solution :

    \[S=\frac{4}{1.3}-\frac{6}{2.4}+\frac{12}{5.7}-\frac{14}{6.8}\]+....  \[=\frac{1+3}{1.3}-\frac{2+4}{2.4}+\frac{5+7}{5.7}-\frac{6+8}{6.8}+....\] \[=\left( \frac{1}{1}+\frac{1}{3} \right)-\left( \frac{1}{2}+\frac{1}{4} \right)-\left( \frac{1}{5}+\frac{1}{7} \right)-\left( \frac{1}{6}+\frac{1}{8} \right)+....\] \[=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....={{\log }_{e}}2\].


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