Question

The value of $\underset{n\to \infty }{lim}\frac{1}{n}\{{\sec }^2\frac{\pi }{4n}+{\sec }^2\frac{2\pi }{4n}+.......+{\sec }^2\frac{n\pi }{4n}\}$ is

• A. ${\log }_{c}2$
• B. $\frac{\pi }{2}$
• C. $\frac{4}{\pi }$
• D. $e$

Answer 1

The correct option is C $\frac{4}{\pi }$

Let,
$L=\underset{n\to \infty }{lim}\frac{1}{n}\{{\sec }^2\frac{\pi }{4n}+{\sec }^2\frac{2\pi }{4n}+.......+{\sec }^2\frac{n\pi }{4n}\}$
$\Rightarrow L=\underset{n\to \infty }{lim}\frac{1}{n}[\sum _{k=1}^n{\sec }^2\left(\frac{k\pi }{4n}\right)]$
Limits as a sum,
$\Rightarrow L=\underset{0}{\overset{1}{\int }}{\sec }^2\left(\frac{\pi x}{4}\right)dx\Rightarrow L=\frac{4}{\pi }[\tan \frac{x\pi }{4}{]}_0^1=\frac{4}{\pi }$