Question

If $S_n={\cot }^{-1}3+{\cot }^{-1}7+{\cot }^{-1}13+{\cot }^{-1}21+...+n$ terms, then

• A. $S_{10}={\tan }^{-1}\frac{5}{6}$
• B. $S_{\infty }=\frac{\pi }{4}$
• C. $S_6={\sin }^{-1}\frac{4}{5}$
• D. $S_{20}={\cot }^{-1}1.1$

Answer 1

Answer: (A), (D)

The correct options are
A $S_{10}={\tan }^{-1}\frac{5}{6}$
D $S_{20}={\cot }^{-1}1.1$

$S_n={\cot }^{-1}3+{\cot }^{-1}7+{\cot }^{-1}13+{\cot }^{-1}21+.....$

$=\sum _{n=1}^n{\cot }^{-1}(1+r+r^2)=\sum co{t}^{-1}(1+r)-co{t}^{-1}r={\cot }^{-1}\left(n+1\right)-\frac{\pi }{4}$

Now $S_{10}={\cot }^{-1}\left(10+1\right)-\frac{\pi }{4}={\cot }^{-1}\left(11\right)-{\cot }^{-1}\left(1\right)$

$={\tan }^{-1}\left(\frac{11-1}{1+11\times 1}\right)={\tan }^{-1}\left(\frac{10}{12}\right)={\tan }^{-1}\left(\frac{5}{6}\right)$

$S_{\infty }=\underset{n\to \infty }{lim}({\cot }^{-1}\left(n+1\right)-\frac{\pi }{4})=\underset{n\to \infty }{lim}({\tan }^{-1}\frac{1}{(n+1)}-\frac{\pi }{4})$

$=0-\frac{\pi }{4}=\frac{-\pi }{4}$

$S_6={\tan }^{-1}\left(\frac{1}{6+1}\right)-\frac{\pi }{4}={\tan }^{-1}\frac{1}{\left(7\right)}-{\tan }^{-1}\left(1\right)$

$={\tan }^{-1}\left(\frac{-7+1}{1+7\times 1}\right)={\tan }^{-1}\left(\frac{6}{8}\right)={\tan }^{-1}\left(\frac{3}{4}\right)$

$={\sin }^{-1}\left(\frac{\left(\frac{3}{4}\right)}{\sqrt{1+{\left(\frac{3}{4}\right)}^2}}\right)={\sin }^{-1}\left(\frac{3}{5}\right)$

$S_{20}={\cot }^{-1}\left(20+1\right)-\frac{\pi }{4}={\cot }^{-1}\left(21\right)-{\cot }^{-1}\left(1\right)$

$={\cot }^{-1}\left(\frac{21+1}{21-1}\right)={\cot }^{-1}\left(\frac{22}{20}\right)$

$={\cot }^{-1}1.1$