Question

$S=ta{n}^{-1}\left(\frac{1}{n^2+n+1}\right)+ta{n}^{-1}\left(\frac{1}{n^2+3n+3}\right)+...$ $ta{n}^{-1}\left(\frac{1}{1+\left(n+19\right)\left(n+20\right)}\right),$ then $tanS$ 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,
$ta{n}^{-1}\frac{1}{1+2}+ta{n}^{-1}+\frac{1}{1+2\times 3}+ta{n}^{-1}\frac{1}{1+3\times 4}+......+$
$ta{n}^{-1}\frac{1}{1+\left(n-1\right)n}+ta{n}^{-1}\frac{1}{1+\left(n\left(n+1\right)\right)}+......+$
$ta{n}^{-1}\frac{1}{1+\left(n+19\right)\left(n+20\right)}=ta{n}^{-1}\frac{n+19}{n+21}$
$\Rightarrow ta{n}^{-1}\frac{n-1}{n+1}+ta{n}^{-1}\frac{1}{1+n\left(n+1\right)}+ta{n}^{-1}\frac{1}{1+\left(n+1\right)\left(n+2\right)}$
$+......+\frac{1}{1+\left(l+\left(n+19\right)\left(n+20\right)\right)}=ta{n}^{-1}\frac{n+19}{n+21}$
$ta{n}^{-1}\frac{1}{1+n\left(n+1\right)}+ta{n}^{-1}\frac{1}{1+\left(n+1\right)\left(n+2\right)}+......+$
$\frac{1}{1+\left(n+19\right)\left(n+20\right)}=ta{n}^{-1}\frac{n+19}{n+21}-ta{n}^{-1}\frac{n+1}{n+1}$
$ta{n}^{-1}\left(\frac{1}{n^2+n+1}\right)+ta{n}^{-1}\left(\frac{1}{n^2+3n+3}\right)+.....+$
$ta{n}^{-1}\frac{1}{1+\left(n+19\right)\left(n+20\right)}$
$=ta{n}^{-1}\left(\frac{\frac{n+19}{n+21}-\frac{n-1}{n+1}}{1+\frac{n+19}{n+21}\times \frac{n-1}{n+1}}\right)$
$=ta{n}^{-1}\frac{20}{n^2+20n+1}=S$
$\therefore ta{n}^{-1}S=\frac{20}{n^2+20n+1}$