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

  • question_answer
    \[1+\left( \frac{1}{2}+\frac{1}{3} \right)\,\frac{1}{4}+\left( \frac{1}{4}+\frac{1}{5} \right)\,\frac{1}{{{4}^{2}}}+\left( \frac{1}{6}+\frac{1}{7} \right)\,\frac{1}{{{4}^{3}}}+....\infty =\]

    A) \[{{\log }_{e}}(2\sqrt{3})\]

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

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

    D) \[{{\log }_{e}}\left( \frac{2}{\sqrt{3}} \right)\]

    Correct Answer: A

    Solution :

    \[S=\left\{ 1+\frac{{{\left( \frac{1}{2} \right)}^{2}}}{2}+\frac{{{\left( \frac{1}{2} \right)}^{4}}}{4}+..... \right\}\]\[+2\left\{ \frac{1}{2}+\frac{{{\left( \frac{1}{2} \right)}^{3}}}{3}+\frac{{{\left( \frac{1}{2} \right)}^{5}}}{5\ }+..... \right\}-1\] \[=1-\frac{1}{2}{{\log }_{e}}\left( 1+\frac{1}{2} \right)\text{ }\left( 1-\frac{1}{2} \right)+{{\log }_{e}}\left( \frac{1+\frac{1}{2}}{1-\frac{1}{2}} \right)-1\] \[=-\frac{1}{2}{{\log }_{e}}\frac{3}{4}+{{\log }_{e}}3={{\log }_{e}}2\sqrt{3}\].


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