JCECE Engineering JCECE Engineering Solved Paper-2015

  • question_answer
    The sum of n terms of the series \[1.4+3.04+5.004+7.0004+...\]is

    A) \[{{n}^{2}}+\frac{4}{9}\left( 1+\frac{1}{{{10}^{n}}} \right)\]         

    B) \[{{n}^{2}}+\frac{4}{9}\left( 1-\frac{1}{{{10}^{n}}} \right)\]

    C) \[n+\frac{4}{9}\left( 1-\frac{1}{{{10}^{n}}} \right)\]         

    D)  None of these

    Correct Answer: B

    Solution :

    \[1.4+3.04+5.004+...\] \[=(1+3+5+...)+(0.4+0.04+0.004+...)\] \[=\frac{n}{2}[2+(n-1)\cdot 2]+\left[ \frac{4}{10}+\frac{4}{{{10}^{2}}}+...\,\,upto\,\,n\,\,terms \right]\]\[={{n}^{2}}+\frac{4}{10}\left[ \frac{1-{{\left( \frac{1}{10} \right)}^{n}}}{1-\frac{1}{10}} \right]\]                 \[={{n}^{2}}+\frac{4}{9}\left( 1-\frac{1}{{{10}^{n}}} \right)\]


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