Question

Let $S_n=\sum _{r=0}^{n-1}{\cos }^{-1}\left(\frac{n^2+r^2+r}{\sqrt{n^4+r^4+2r^3+2n^2{r}^2+2n^{2}r+n^2+r^2}}\right)$. Then the value of $S_{100}$, is

• A. $\frac{\pi }{12}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (D)

$S_n=\sum _{r=0}^{n-1}{\tan }^{-1}\left(\frac{n}{n^2+r(r+1)}\right)$
$S_n=\sum _{r=0}^{n-1}{\tan }^{-1}\left(\frac{\frac{1}{n}}{1+\frac{r+1}{n}\cdot \frac{r}{n}}\right)$
$S_n=\sum _{r=0}^{n-1}{\tan }^{-1}\left(\frac{\frac{r+1}{n}-\frac{r}{n}}{1+\frac{r+1}{n}\cdot \frac{r}{n}}\right)$
$S_n=\sum _{r=0}^{n-1}{\tan }^{-1}\left(\frac{r+1}{n}\right)-{\tan }^{-1}\frac{r}{n}$
$S_n=\left({\tan }^{-1}\frac{1}{n}-0\right)+\left({\tan }^{-1}\frac{2}{n}-{\tan }^{-1}\frac{1}{n}\right)+\left({\tan }^{-1}1-{\tan }^{-1}\frac{n-1}{n}\right)$
$\therefore S_n=S_{100}=\frac{\pi }{4}$