Question

Let $S_n=\displaystyle\sum_{r=0}^{n-1} \cos ^{-1}\left(\frac{n^2+r^2+r}{\sqrt{n^4+r^4+2 r^3+2 n^2 r^2+2 n^2 r+n^2+r^2}}\right)$. 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)

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