Question

$S=tan^{-1}\left(\frac{1}{n^{2}+n+1}\right)+tan^{-1}\left(\frac{1}{n^{2}+3n+3}\right)+...$ $tan^{-1} \left(\frac{1}{1+\left(n + 19\right)\left(n + 20\right)}\right),$ then $tan\, S$ is equal to :

• A. $\frac{20}{401 + 20n}$
• B. $\frac{n}{n^{2} + 20n+1}$
• C. $\frac{20}{n^{2} + 20n+1}$
• D. $\frac{n}{401 + 20n}$

Answer 1

Answer: (C)

We know that,
$tan^{-1} \frac{1}{1+2}+tan^{-1}+\frac{1}{1+2\times3}+tan^{-1} \frac{1}{1+3\times4}+...... +$
$tan^{-1} \frac{1}{1+\left(n-1\right)n}+tan^{-1} \frac{1}{1+\left(n\left(n+1\right)\right)} + ......+$
$tan^{-1} \frac{1}{1+\left(n+19\right)\left(n+20\right)}=tan^{-1} \frac{n+19}{n+21}$
$\Rightarrow tan^{-1} \frac{n-1}{n+1}+tan^{-1} \frac{1}{1+n\left(n+1\right)}+tan^{-1} \frac{1}{1+\left(n+1\right)\left(n+2\right)}$
$+......+ \frac{1}{1+\left(l + \left(n + 19\right) \left(n + 20\right)\right)}=tan^{-1} \frac{n+19}{n+21}$
$tan^{-1} \frac{1}{1+n\left(n+1\right)}+tan^{-1} \frac{1}{1+\left(n+1\right)\left(n+2\right)}+......+$
$\frac{1}{1 + \left(n+19\right) \left(n + 20\right)}=tan^{-1} \frac{n+19}{n+21}-tan^{-1} \frac{n+1}{n+1}$
$tan^{-1} \left(\frac{1}{n^{2}+n+1}\right)+tan^{-1} \left(\frac{1}{n^{2}+3n+3}\right)+.....+$
$tan^{-1} \frac{1}{1+\left(n+19\right)\left(n+20\right)}$
$=tan^{-1}\left(\frac{\frac{n+19}{n+21}-\frac{n-1}{n+1}}{1+\frac{n+19}{n+21}\times\frac{n-1}{n+1}}\right)$
$=tan^{-1} \frac{20}{n^{2}+20n+1}=S$
$\therefore tan^{-1}\, S=\frac{20}{n^{2}+20n+1}$