MCQ
$\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\left( {n + 1} \right)}^{1/3}}}}{{{n^{4/3}}}} + \frac{{{{\left( {n + 2} \right)}^{1/3}}}}{{{n^{4/3}}}} + .... + \frac{{{{\left( {2n} \right)}^{1/3}}}}{{{n^{4/3}}}}} \right)$ is equal to
  • $\frac{3}{4}{\left( 2 \right)^{4/3}} - \frac{3}{4}$
  • B
    $\frac{4}{3}{\left( 2 \right)^{3/4}}$
  • C
    $\frac{3}{4}{\left( 2 \right)^{4/3}} - \frac{4}{3}$
  • D
    $\frac{4}{3}{\left( 2 \right)^{4/3}}$

Answer

Correct option: A.
$\frac{3}{4}{\left( 2 \right)^{4/3}} - \frac{3}{4}$
a
$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {\frac{1}{n}} {\left( {\frac{{n + r}}{n}} \right)^{1/3}}$

$ = \int\limits_0^1 {{{\left( {1 + x} \right)}^{1/3}}} dx = \frac{3}{4}\left( {{2^{4/3}} - 1} \right)$

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