Question

The value of $\underset{n\to \infty }{lim}(\frac{\sqrt{n}}{{(3+4\sqrt{n})}^2}+\frac{\sqrt{n}}{\sqrt{2}{(3\sqrt{2}+4\sqrt{n})}^2}+\frac{\sqrt{n}}{\sqrt{3}{(3\sqrt{3}+4\sqrt{n})}^2}+\cdots +\frac{1}{49n})$
is of the form $\frac{1}{p},$ where $p\in N.$ Then possible factors of $p$ is/are

• A. $7$
• B. $2$
• C. $3$
• D. $5$

Answer 1

Answer: (A), (B)

$2$

Let $L=\underset{n\to \infty }{lim}(\frac{\sqrt{n}}{{(3+4\sqrt{n})}^2}+\frac{\sqrt{n}}{\sqrt{2}{(3\sqrt{2}+4\sqrt{n})}^2}+\cdots +\frac{\sqrt{n}}{\sqrt{n}{(3\sqrt{n}+4\sqrt{n})}^2})$
The above expression can be written as
$L=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{\sqrt{n}}{\sqrt{r}{(3\sqrt{r}+4\sqrt{n})}^2}=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{n\sqrt{\frac{r}{n}}{(3\sqrt{\frac{r}{n}}+4)}^2}=\underset{0}{\overset{1}{\int }}\frac{dx}{\sqrt{x}{(3\sqrt{x}+4)}^2}$
Putting $(3\sqrt{x}+4)=t\Rightarrow \frac{3}{2\sqrt{x}}dx=dt$
When $x\to 0,(3\sqrt{x}+4)\to 4$
When $x\to 1,(3\sqrt{x}+4)\to 7$
Hence, $\frac{1}{p}=\frac{2}{3}\underset{4}{\overset{7}{\int }}\frac{dt}{t^2}$
$\Rightarrow \frac{1}{p}=\frac{2}{3}{[-\frac{1}{t}]}_4^7=\frac{2}{3}(-\frac{1}{7}+\frac{1}{4})=\frac{1}{14}$
Clearly, the value of $p$ is $14$ and the factors are $1,2,7,14.$