J & K CET Engineering J and K - CET Engineering Solved Paper-2012

  • question_answer
    The sum of n terms of the series \[\frac{3}{{{1}^{2}}{{.2}^{2}}}+\frac{5}{{{2}^{2}}{{.3}^{2}}}+\frac{7}{{{3}^{2}}{{.4}^{2}}}+.....\] is equal to

    A)  \[\frac{{{n}^{2}}-2n}{{{(n+1)}^{2}}}\]

    B)  \[\frac{{{n}^{2}}-2}{{{(n+1)}^{2}}}\]

    C)  \[\frac{{{n}^{2}}+2n}{{{(n+1)}^{2}}}\]

    D)  \[\frac{{{n}^{2}}+2}{{{(n+1)}^{2}}}\]

    Correct Answer: C

    Solution :

    Let \[{{S}_{n}}=\frac{3}{{{1}^{2}}{{.2}^{2}}}+\frac{5}{{{2}^{2}}{{.3}^{2}}}+....+\frac{(2n+1)}{{{n}^{2}}.{{(n+1)}^{2}}}\] \[=\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)+\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)\] \[+......+\left( \frac{1}{{{n}^{2}}}-\frac{1}{{{(n+1)}^{2}}} \right)\] \[=\frac{1}{1}-\frac{1}{{{(n+1)}^{2}}}\] \[=\frac{{{n}^{2}}+1+2n-1}{{{(n+1)}^{2}}}\] \[=\frac{{{n}^{2}}+2n}{{{(n+1)}^{2}}}\]


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