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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