Question

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

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

Answer 1

Answer: (C)

$\frac{1}{10}$

$\underset{n\to \infty }{lim}\sum _{r=1}^{4n}\frac{\sqrt{n}}{\sqrt{r}{(3\sqrt{r}+4\sqrt{n})}^2}$
$T_r=\frac{1}{\sqrt{\frac{r}{n}}n{(3\sqrt{\frac{r}{n}}+4)}^2}$
$\Rightarrow S=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^{4n}\frac{1}{\sqrt{\frac{r}{n}}{(3\sqrt{\frac{r}{n}}+4)}^2}$
Put $\sqrt{\frac{r}{n}}=x$
$\Rightarrow S={\int }_0^4\frac{dx}{\sqrt{x}{(3\sqrt{x}+4)}^2}$

Put $3\sqrt{x}+4=t$
$\Rightarrow \frac{3}{2}.\frac{1}{\sqrt{x}}dx=dt$
$S=\frac{2}{3}{\int }_4^{10}\frac{dt}{t^2}$
$=\frac{2}{3}{\left[\frac{-1}{t}\right]}_4^{10}$

$=\frac{2}{3}{\left[\frac{1}{t}\right]}_{10}^4$
$=\frac{1}{10}$