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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2006

  • question_answer
    If  \[{{S}_{n}}=\frac{1}{6.11}+\frac{1}{11.16}+\frac{1}{16.21}+.....\] to n terms, then \[6{{S}_{n}}\], equals :

    A)  \[\frac{5n-4}{5n+6}\]                   

    B)  \[\frac{n}{(5n+6)}\]

    C)  \[\frac{2n-1}{5n+6}\]                   

    D)  \[\frac{1}{(5n+6)}\]

    Correct Answer: B

    Solution :

    Given that \[{{S}_{n}}=\frac{1}{6.11}+\frac{1}{11.16}+\frac{1}{16.21}+....+n\] terms \[=\frac{1}{5}\left( \frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{5n+1}-\frac{1}{5n+6} \right)\]                 \[=\frac{1}{5}\left( \frac{1}{6}-\frac{1}{5n+6} \right)\]                 \[=\frac{1}{5}\left( \frac{5n+6-6}{6(5n+6)} \right)\]                 \[=\frac{n}{6(5n+6)}\] \[\Rightarrow \]               \[6{{S}_{n}}=\frac{n}{5n+6}\]


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