Question

The value of $\underset{n\to \infty }{lim}(\frac{1}{\sqrt{4n^2-1}}+\frac{1}{\sqrt{4n^2-4}}+....+\frac{1}{\sqrt{4n^2-2n}})$ is.

• A. $\frac{1}{4}$
• B. $\frac{\pi }{12}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{6}$

Answer 1

Answer: (D)

$\frac{\pi }{6}$

$I=\underset{n\to \infty }{lim}(\frac{1}{\sqrt{{4n}^2-1}}+\frac{1}{\sqrt{{4n}^2-4}}+\frac{1}{\sqrt{{4n}^2-9}}........\frac{1}{\sqrt{{4n}^2-n^2}})=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{\sqrt{{4n}^2-r^2}}=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^n\frac{1}{\sqrt{4-\frac{r^2}{n^2}}}$

Here by general method to solve such problems we assume $\frac{r}{n}=x$ and $\frac{1}{n}=dx$

$={\int }_0^1\frac{1}{\sqrt{4-x^2}}dx={\left({\sin }^{-1}\left(\frac{x}{2}\right)\right)}_0^1={\sin }^{-1}\left(\frac{1}{2}\right)-{\sin }^{-1}\left(\frac{0}{2}\right)=\frac{\pi }{6}-0=\frac{\pi }{6}$

Hence, option $D$ is correct.