Question

The value of $\underset{n\to \infty }{\mathrm{Lim}}\left(\frac{1-\cos \frac{2}{n}}{1+\cos \frac{2}{n}}\right)\left(\sum _{r=1}^{2n}\frac{n}{\sqrt{8n^2-r^2}}\right)\left(\sum _{r=1}^{3n}\frac{n}{\sqrt{18n^2-r^2}}\right)$ is equal to

• A. $\frac{{\pi }^2}{4}$
• B. $\frac{{\pi }^2}{8}$
• C. $\frac{{\pi }^2}{16}$
• D. $\frac{3{\pi }^2}{16}$

Answer 1

Answer: (C)

$\underset{n\to \infty }{\mathrm{Lim}}\left(\frac{1-\cos \frac{2}{n}}{1+\cos \frac{2}{n}}\right)\left(\sum _{r=1}^{2n}\frac{n}{\sqrt{8n^2-r^2}}\right)\left(\sum _{r=1}^{3n}\frac{n}{\sqrt{18n^2-r^2}}\right)$
$=\underset{n\to \infty }{\mathrm{Lim}}\frac{1}{2}\cdot \left(\frac{1-\cos \frac{2}{n}}{\frac{4}{n^2}}\right)\cdot 4\cdot \left(\sum _{r=1}^{2n}\frac{n}{n\cdot n\sqrt{8-\frac{r^2}{n^2}}}\right)\left(\sum _{r=1}^{3n}\frac{n}{n\cdot n\sqrt{18-\frac{r^2}{n^2}}}\right)$
$=\frac{1}{2}\times \frac{1}{2}\times 4\left(\underset{0}{\overset{2}{\int }}\frac{dx}{\sqrt{8-x^2}}\right)\left(\underset{0}{\overset{3}{\int }}\frac{dx}{\sqrt{18-x^2}}\right)={\left({\sin }^{-1}\frac{x}{2\sqrt{2}}\right)}_0^2{\left({\sin }^{-1}\frac{x}{3\sqrt{2}}\right)}_0^3=\frac{\pi }{4}\cdot \frac{\pi }{4}=\frac{{\pi }^2}{16}$