Question

If $S_n={\cot }^{-1}\left(3\right)+{\cot }^{-1}\left(7\right)+{\cot }^{-1}\left(13\right)+{\cot }^{-1}\left(21\right)+\dots n\text{terms}$, then

• A. $S_{10}={\tan }^{-1}\frac{5}{6}$
• B. $S_{\infty }=\frac{\pi }{4}$
• C. $S_6={\sin }^{-1}\frac{3}{5}$
• D. $S_{20}={\cot }^{-1}\left(1.1\right)$

Answer 1

Answer: (A), (B), (C), (D)

$S_{20}={\cot }^{-1}\left(1.1\right)$

Let $t_r$ denote the $r$th term of the series $3,7,13,21,...$ and
$S=3+7+13+21+...+t_n$
$-S=0-3-7-13-21-...-t_{n-1}-t_n$
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$0=3+4+6+8+...+2n-t_n$
$\Rightarrow t_n=3+4+6+...+2n=1+2\times \frac{1}{2}n(n+1)$
$=n^2+n+1$
Let $T_r={\cot }^{-1}(r^2+r+1)$
$={\tan }^{-1}\left(\frac{1}{r^2+r+1}\right)$

$={\tan }^{-1}\left(\frac{r+1-r}{1+r(r+1)}\right)$
$={\tan }^{-1}(r+1)-{\tan }^{-1}r$
Thus, the sum of first $n$ terms of the given series is
${\sum }_{r=1}^n\left[{\tan }^{-1}\right(r+1)-{\tan }^{-1}r]$
$={\tan }^{-1}(n+1)-{\tan }^{-1}\left(1\right)$
$={\tan }^{-1}\left[\frac{n+1-1}{1+1(n+1)}\right]={\tan }^{-1}\left(\frac{n}{n+2}\right)$
$={\tan }^{-1}\left(\frac{1}{1+\frac{2}{n}}\right)$
$S_{\infty }=\underset{n\to \infty }{lim}{\tan }^{-1}\left(\frac{1}{1+\frac{2}{n}}\right)=\frac{\pi }{4},$

$S_{10}={\tan }^{-1}\frac{10}{12}={\tan }^{-1}\frac{5}{6}$

$S_6={\tan }^{-1}\frac{3}{4}={\sin }^{-1}\frac{3}{5}$

$S_{20}={\tan }^{-1}\frac{10}{11}={\cot }^{-1}1.1$