Question

If $S_n$ denotes the sum to $n$ terms of the series
${\cot }^{-1}\frac{7}{4}+{\cot }^{-1}\frac{19}{4}+{\cot }^{-1}\frac{39}{4}+{\cot }^{-1}\frac{67}{4}+....$ then

• A. $S_n={\tan }^{-1}\frac{n}{2n+5}$
• B. $S_n={\cot }^{-1}\frac{n+5}{2n}$
• C. $S_n={\tan }^{-1}\frac{4n}{2n+5}$
• D. ${lim}_{n\to \infty }{S}_n={\cot }^{-1}\frac{1}{2}$

Answer 1

The correct option is C $S_n={\tan }^{-1}\frac{4n}{2n+5}$

${\cot }^{-1}\frac{7}{4}+{\cot }^{-1}\frac{19}{4}+{\cot }^{-1}\frac{39}{4}+........$

$={\cot }^{-1}\left(\frac{4+3}{4}\right)+{\cot }^{-1}\left(\frac{16+3}{4}\right)+{\cot }^{-1}\left(\frac{30+3}{4}\right)+......$

$={\cot }^{-1}(1+\frac{3}{4})+{\cot }^{-1}(4+\frac{3}{4})+{\cot }^{-1}(9+\frac{3}{4})+....$

$={\sum }_{r=1}^n{\cot }^{-1}(r^2+\frac{3}{4})$

$={\sum }_{r=1}^r{\tan }^{-1}\left(\frac{1}{r^2+3/4}\right)=\sum {\tan }^{-1}\left(\frac{1}{1+(r^2-1/4)}\right)$

$=\sum {\tan }^{-1}\left(\frac{1}{1+(r+1/2)(r-1/2)}\right)=\sum {\tan }^{-1}\left(\frac{(r+1/2)-(r-1/2)}{1+(r+1/2)(r-1/2)}\right)$

$={\sum }_{r=1}^n{\tan }^{-1}(r+1/2)-{\tan }^{-1}(r-1/2)$

$={\tan }^{-1}(n+\frac{1}{2})$

$={\tan }^{-1}(n+\frac{1}{2})-\left(\frac{1}{2}\right)$

$={\tan }^{-1}\left(\frac{n+1\frac{1}{2}-\frac{1}{2}}{1+\frac{1}{2}(n+\frac{1}{2})}\right)={\tan }^{-1}\left(\frac{n}{1+\frac{n}{2}+\frac{1}{4}}\right)$

$={\tan }^{-1}\left(\frac{n}{\frac{n}{2}+\frac{5}{4}}\right)={\tan }^{-1}\left(\frac{4n}{5+2n}\right)$