JCECE Engineering JCECE Engineering Solved Paper-2003

  • question_answer
    The sum of\[\frac{1}{2}+\frac{1}{3}+\frac{1}{{{2}^{3}}}+\frac{1}{5}\cdot \frac{1}{{{2}^{5}}}+...\infty \]is:

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

    B) \[{{\log }_{e}}\sqrt{3}\]

    C) \[{{\log }_{e}}\sqrt{\frac{7}{2}}\]                             

    D)  \[{{\log }_{e}}3\]

    Correct Answer: B

    Solution :

    Let          \[S=\frac{1}{2}+\frac{1}{3}\cdot \frac{1}{{{2}^{3}}}+\frac{1}{5}\cdot \frac{1}{{{2}^{5}}}+...\]                    \[=\frac{1}{2}\log \left( \frac{1+\frac{1}{2}}{1-\frac{1}{2}} \right)\]                 \[\left[ \because \,\,\log \left( \frac{1+x}{1-x} \right)=2\left( x+\frac{{{x}^{3}}}{3}+... \right) \right]\]                 \[=\frac{1}{2}\log \left( \frac{3}{1} \right)=\log {{3}^{1/2}}\]                 \[=\log \sqrt{3}\]


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