JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer \[\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{{{(n+1)}^{{}^{1}/{}_{3}}}}{{{n}^{{}^{4}/{}_{3}}}}+\frac{{{(n+2)}^{{}^{1}/{}_{3}}}}{{{n}^{{}^{4}/{}_{3}}}}+......+\frac{{{(2n)}^{{}^{1}/{}_{3}}}}{{{n}^{{}^{4}/{}_{3}}}} \right)\] is equal to: [JEE Main 10-4-2019 Morning]

    A) \[\frac{4}{3}{{(2)}^{{}^{4}/{}_{3}}}\]             

    B) \[\frac{3}{4}{{(2)}^{{}^{4}/{}_{3}}}-\frac{4}{3}\]

    C) \[\frac{3}{4}{{(2)}^{{}^{4}/{}_{3}}}-\frac{3}{4}\]    

    D) \[\frac{4}{3}{{(2)}^{{}^{3}/{}_{4}}}\]

    Correct Answer: C

    Solution :

                \[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{\frac{1}{n}}{{\left( \frac{n+r}{n} \right)}^{1/3}}\]           \[=\int\limits_{0}^{1}{{{\left( 1+x \right)}^{1/3}}}dx=\frac{3}{4}\left( {{2}^{4/3}}-1 \right)\]


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