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JEE Mains · Maths · STD 11 - 12. limits
\(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\left( {n + 1} \right)}^{1/3}}}}{{{n^{4/3}}}} + \frac{{{{\left( {n + 2} \right)}^{1/3}}}}{{{n^{4/3}}}} + .... + \frac{{{{\left( {2n} \right)}^{1/3}}}}{{{n^{4/3}}}}} \right)\) is equal to
- A \(\frac{3}{4}{\left( 2 \right)^{4/3}} - \frac{3}{4}\)
- 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 & Solution
Correct Answer
(A) \(\frac{3}{4}{\left( 2 \right)^{4/3}} - \frac{3}{4}\)
Step-by-step Solution
Detailed explanation
\(\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {\frac{1}{n}} {\left( {\frac{{n + r}}{n}} \right)^{1/3}}\) \( = \int\limits_0^1 {{{\left( {1 + x} \right)}^{1/3}}} dx = \frac{3}{4}\left( {{2^{4/3}} - 1} \right)\)
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