Q120 / 200 · Manipal (MET) 2011 · Mathematics
$\lim _{n \rightarrow \infty} \frac{1}{2}\left[\sec ^2 \frac{\pi}{4 n}+\sec ^2 \frac{2 \pi}{4 n}+\ldots+\sec ^2 \frac{n \pi}{4 n}\right]$ is equal to
- Correct answerA$\frac{4}{\pi}$
- B$\frac{2}{\pi}$
- C$\frac{3}{\pi}$
- D$\frac{5}{\pi}$
Solution
$\begin{aligned} \lim _{n \rightarrow \infty} \frac{1}{n} & {\left[\sec ^2 \frac{\pi}{4 n}+\sec ^2 \frac{2 \pi}{4 n}+\ldots+\sec ^2 \frac{n \pi}{4 n}\right] } \\ & =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \sec ^2 \frac{r \pi}{4 n} \\ & =\int_0^1 \sec ^2 \frac{\pi x}{4} d x \\ & =\frac{4}{\pi}\left[\tan \frac{\pi x}{4}\right]_0^1=\frac{4}{\pi}\end{aligned}$
Hence, the correct answer is option: 'A'.