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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 1

  • question_answer
    \[\underset{n\to \infty }{\mathop{\lim }}\,\frac{(1+2+3+...+n({{1}^{3}}+{{2}^{3}}+{{3}^{3}}+...+{{n}^{3}})}{{{({{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{n}^{2}})}^{2}}}=\]

    A) \[\frac{3}{8}\]              

    B) \[\frac{5}{8}\]

    C) \[\frac{7}{8}\]              

    D) \[\frac{9}{8}\]

    Correct Answer: B

    Solution :

    [b]: \[\underset{n\to \infty }{\mathop{\lim }}\,\frac{n(n+1)}{2}.\frac{{{n}^{2}}{{(n+1)}^{2}}}{4}.\frac{36}{{{n}^{2}}{{(n+1)}^{2}}{{(2n+1)}^{2}}}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{9\left( 1+\frac{1}{n} \right)}{{{\left( 2+\frac{1}{n} \right)}^{2}}}=\frac{9}{8}\]


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