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12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    \[\underset{n\to \infty }{\mathop{\lim }}\,\frac{1+{{2}^{4}}+{{3}^{4}}+....+{{n}^{4}}}{{{n}^{5}}}\]\[-\underset{n\to \infty }{\mathop{\lim }}\,\frac{1+{{2}^{3}}+{{3}^{3}}+....+{{n}^{3}}}{{{n}^{5}}}=\] [AIEEE 2003]

    A) \[\frac{1}{30}\]                   

    B) Zero

    C) \[\frac{1}{4}\]                      

    D) \[\frac{1}{5}\]

    Correct Answer: D

    Solution :

    • \[\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\left\{ {{\left( \frac{1}{n} \right)}^{4}}+{{\left( \frac{2}{n} \right)}^{4}}+{{\left( \frac{3}{n} \right)}^{4}}+...+{{\left( \frac{n}{n} \right)}^{4}} \right\}-\]\[\] \[\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\left\{ \left( \frac{1}{{{n}^{4}}} \right)+\left( \frac{{{2}^{3}}}{{{n}^{4}}} \right)+......+\left( \frac{{{n}^{3}}}{{{n}^{4}}} \right) \right\}\]                   
    • \[=\int_{0}^{1}{{{(x)}^{4}}dx-0=\left[ \frac{{{x}^{5}}}{5} \right]_{0}^{1}=\frac{1}{5}.}\]


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