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JEE Main & Advanced JEE Main Paper (Held On 9 April 2017)

  • question_answer
    Let \[{{S}_{n}}=\frac{1}{{{1}^{3}}}+\frac{1+2}{{{1}^{3}}+{{2}^{3}}}+\frac{1+2+3}{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}+...+\frac{1+2+....,\,+n}{{{1}^{3}}+{{2}^{3}}+...+{{n}^{3}}}\].  If \[100\,{{S}_{n}}=n,\] then n is equal to: [JEE Online 09-04-2017]

    A)  200                                       

    B)  199

    C)  99                                         

    D)  19

    Correct Answer: B

    Solution :

                    \[{{T}_{n}}=\frac{\frac{n+(n+1)}{2}}{{{\left( \frac{n+(n+1)}{2} \right)}^{2}}}\] \[{{T}_{n}}=\frac{2}{n(n+1)}\] \[{{S}_{n}}=2\sum\limits_{n=1}^{n}{\left( \frac{1}{n}-\frac{1}{n+1} \right)}\] \[=2\,\left\{ \begin{matrix}    1-\frac{1}{2}  \\    \frac{1}{2}-\frac{1}{3}  \\ \end{matrix} \right.\]\[\left. \frac{1}{n}-\,\frac{1}{n+1} \right\}\] \[=2\left\{ 1-\frac{1}{n+1} \right\}\] \[\] \[100\times \frac{2n}{n+1}=n\] \[n+1=200\] \[n=199\]


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