Question

The value of $\underset{n\to \infty }{lim}\sum _{r=1}^{r=4n}\frac{\sqrt{n}}{\sqrt{r}{(3\sqrt{r}+\sqrt{n})}^2}$ is equal to

• A. $\frac{1}{35}$
• B. $\frac{1}{14}$
• C. $\frac{1}{10}$
• D. $\frac{1}{5}$

Answer 1

The correct option is C $\frac{1}{10}$

Let $I=\frac{lim}{n\to \infty }\sum _{r=1}^{r=4n}\frac{\sqrt{n}}{\sqrt{r}{(3\sqrt{r}+4\sqrt{n})}^2}$
$=\frac{lim}{n\to \infty }\frac{1}{n}\sum _{r=1}^{4n}\frac{1}{{(3\sqrt{\frac{r}{n}}+4)}^2\sqrt{\frac{r}{n}}}$
$I={\int }_0^4\frac{dx}{\sqrt{x}{(3\sqrt{x}+4)}^2}$
Put $3\sqrt{x}+4=t\Rightarrow \frac{3}{2\sqrt{x}}dx=dt.$
$I=\frac{2}{3}{\int }_4^{10}\frac{dt}{t^2}=\frac{2}{3}{\left[\frac{1}{t}\right]}_{10}^4$
$=\frac{2}{3}.\frac{6}{40}=\frac{1}{10.}$