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Manipal Engineering Manipal Engineering Solved Paper-2008

  • question_answer
    Let\[{{S}_{n}}=\frac{1}{{{1}^{3}}}+\frac{1}{{{1}^{3}}}+\frac{2}{{{2}^{3}}}+...+\frac{1+2+...+n}{{{1}^{3}}+{{2}^{3}}+...+n}\] \[n=1,2,3,...\]. Then,\[{{S}_{n}}\]is not greater than

    A) \[\frac{1}{2}\]                                   

    B)  1

    C)  2                            

    D)         4

    Correct Answer: C

    Solution :

    Given,                 \[{{S}_{n}}=\frac{1}{{{1}^{3}}}+\frac{1+2}{{{1}^{3}}+{{2}^{3}}}\]                                 \[+...+\frac{1+2+3+...+n}{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+...+{{n}^{3}}}\] Now,     \[{{T}_{n}}=\frac{1+2+3+...+n}{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+...+{{n}^{3}}}=\frac{\Sigma n}{\Sigma {{n}^{3}}}\]                 \[=\frac{n(n+1)/2}{{{\{n(n+1)/2\}}^{2}}}\]                 \[=\frac{2}{n(n+1)}\]                 \[=2\left( \frac{1}{n}-\frac{1}{n+1} \right)\] \[\therefore \]  \[{{S}_{n}}={{T}_{1}}+{{T}_{2}}+...+{{T}_{n}}\]                 \[=2\left( \frac{1}{1}-\frac{1}{2} \right)+2\left( \frac{1}{2}-\frac{1}{3} \right)+...\]                                                 \[+2\left( \frac{1}{n}-\frac{1}{n+1} \right)\]                 \[=2\left( 1-\frac{1}{n+1} \right)\]                 \[=2-\frac{2}{n+1}\le 2\]                               \[\left( \because \,\,\frac{2}{n+1}\le 1 \right)\]


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