Question

$\stackrel{lim}{n\to \infty }(\frac{1}{n^2}se{c}^2\frac{1}{n^2}+\frac{2}{n^2}se{c}^2\frac{4}{n^2}+...+\frac{n}{n^2}se{c}^{2}1)equalsto$

• A. $\frac{1}{2}tan1$
• B. $tan1$
• C. $\frac{1}{2}cosec1$
• D. $\frac{1}{2}sec1$

Answer 1

The correct option is A $\frac{1}{2}tan1$

$=\stackrel{lim}{n\to \infty }(\frac{1}{n^2}se{c}^2\frac{n}{n^2}+\frac{2}{n^2}se{c}^2\frac{4}{n^2}+....+\frac{n}{n^2}se{c}^{2}1)$
$=\stackrel{lim}{n\to \infty }\frac{1}{n}(\frac{1}{n}se{c}^2\frac{1}{n^2}+\frac{2}{n}se{c}^2\frac{4}{n^2}+...+\frac{n}{n}se{c}^{2}1)$
$=\stackrel{lim}{n\to \infty }\frac{1}{n}{\sum }_{r=1}^n\left(\frac{r}{n}\right)se{c}^2{\left(\frac{r}{n}\right)}^2$
$\therefore A={\int }_0^{1}xse{c}^2\left(x^2\right)dx$
$Put{x}^2=t$
$\Rightarrow 2xdx=dt\Rightarrow xdx=\frac{dt}{2}$
$\therefore A=\frac{1}{2}{\int }_0^{1}se{c}^{2}tdt=\frac{1}{2}tan1$