CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    The sum of the infinite series\[\frac{1}{2}\left( \frac{1}{3}+\frac{1}{4} \right)-\frac{1}{4}\left( \frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{2}}} \right)+\frac{1}{6}\left( \frac{1}{{{3}^{3}}}+\frac{1}{{{4}^{3}}} \right)-....\]is equal to

    A)  \[\frac{1}{2}\log 2\]      

    B)         \[\log \frac{3}{5}\]

    C)  \[\log \frac{5}{3}\]         

    D)         \[\frac{1}{2}\log \frac{5}{3}\]

    E)  \[\frac{1}{2}\log \frac{3}{5}\]

    Correct Answer: D

    Solution :

    \[\frac{1}{2}\left( \frac{1}{3}+\frac{1}{4} \right)-\frac{1}{4}\left( \frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{2}}} \right)+\frac{1}{6}\left( \frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{3}}} \right)-....\] \[=\frac{1}{2}\left[ \frac{1}{3}-\frac{1}{2}\left( \frac{1}{{{3}^{2}}} \right)+\frac{1}{3}\left( \frac{1}{{{3}^{3}}} \right)-.... \right]\]                 \[+\frac{1}{2}\left[ \frac{1}{4}-\frac{1}{2}\left( \frac{1}{{{4}^{2}}} \right)+\frac{1}{3}\left( \frac{1}{{{4}^{3}}} \right)-..... \right]\] \[=\frac{1}{2}\left[ \log \left( 1+\frac{1}{3} \right) \right]+\frac{1}{2}\left[ \log \left( 1+\frac{1}{4} \right) \right]\]                 \[=\frac{1}{2}\left[ \log \left( \frac{4}{3} \right)\times \left( \frac{5}{4} \right) \right]\]                 \[=\frac{1}{2}\log \left( \frac{5}{3} \right)\]        


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