Question

$\underset{n\to \infty }{lim}(\frac{(n+1{)}^{1/3}}{n^{4/3}}+\frac{(n+2{)}^{1/3}}{n^{4/3}}+\cdots +\frac{(2n{)}^{1/3}}{n^{4/3}})$ is equal to

• A. $\frac{3}{4}(2{)}^{4/3}-\frac{3}{4}$
• B. $\frac{4}{3}(2{)}^{4/3}$
• C. $\frac{3}{4}(2{)}^{4/3}-\frac{4}{3}$
• D. $\frac{4}{3}(2{)}^{3/4}$

Answer 1

The correct option is A $\frac{3}{4}(2{)}^{4/3}-\frac{3}{4}$

$\underset{n\to \infty }{lim}\frac{1}{n}({(1+\frac{1}{n})}^{1/3}+{(1+\frac{2}{n})}^{1/3}+\cdots +{(1+\frac{n}{n})}^{1/3})$
$=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{n}{(1+\frac{r}{n})}^{1/3}$
$\frac{r}{n}\to x,\frac{1}{n}\to dx$
$=\underset{0}{\overset{1}{\int }}(1+x{)}^{1/3}dx=\frac{3}{4}(2^{4/3}-1)$