Question

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

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

Answer 1

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

$T_r=\frac{\sqrt{n}}{\sqrt{r}(3\sqrt{r}+4\sqrt{n}{)}^2}\Rightarrow T_r=\frac{1}{\sqrt{\frac{r}{n}}n{(3\sqrt{\frac{r}{n}}+4)}^2}$
$\therefore S=\underset{n\to \infty }{lim}\frac{1}{n}\sum _1^{4n}\frac{1}{\sqrt{\frac{r}{n}}{(3\sqrt{\frac{r}{n}}+4)}^2}$
$\Rightarrow S=\underset{0}{\overset{4}{\int }}\frac{dx}{\sqrt{x}(3\sqrt{x}+4{)}^2}$

Putting $3\sqrt{x}+4=t$
$\Rightarrow \frac{3}{2\sqrt{x}}dx=dt$
$\therefore S=\frac{2}{3}\underset{4}{\overset{10}{\int }}\frac{dt}{t^2}=\frac{2}{3}{\left[\frac{1}{t}\right]}_{10}^4=\frac{1}{10}$