S=tan ^-1 ( 1 n^2+n+1 )+tan ^-1 ( 1 n^2+3 n+3 )+… +tan ^-1 ( 1 1+(n+19)(n+20) ), then tan S is equal to :

We know that, $$\begin{array}{r} \tan ^{-1} \frac{1}{1+2}+\tan ^{-1} \frac{1}{1+2 \times 3}+\tan ^{-1} \frac{1}{1+3 \times 4}+\ldots \ldots+ \\ \tan ^{-1} \frac{1}{1+(n-1) n}+\tan ^{-1} \frac{1}{1+n(n+1)}+\ldots \ldots+ \\ \tan ^{-1} \frac{1}{1+(n+19)(n+20)}=\tan ^{-1} \frac{n+19}{n+21} \\ \Rightarrow \tan ^{-1} \frac{n-1}{n+1}+\tan ^{-1} \frac{1}{1+n(n+1)}+\tan ^{-1} \frac{1}{1+(n+1)(n+2)} \\ +\ldots \ldots+\frac{1}{1+(n+19)(n+20)}=\tan ^{-1} \frac{n+19}{n+21} \\ \tan ^{-1} \frac{1}{1+n(n+1)}+\tan ^{-1} \frac{1}{1+(n+1)(n+2)}+\ldots \ldots+ \\ \frac{1}{1+(n+19)(n+20)}=\tan ^{-1} \frac{n+19}{n+21}-\tan ^{-1} \frac{n-1}{n+1} \end{array}$$tan−11+21​+tan−11+2×31​+tan−11+3×41​+……+tan−11+(n−1)n1​+tan−11+n(n+1)1​+……+tan−11+(n+19)(n+20)1​=tan−1n+21n+19​⇒tan−1n+1n−1​+tan−11+n(n+1)1​+tan−11+(n+1)(n+2)1​+……+1+(n+19)(n+20)1​=tan−1n+21n+19​tan−11+n(n+1)1​+tan−11+(n+1)(n+2)1​+……+1+(n+19)(n+20)1​=tan−1n+21n+19​−tan−1n+1n−1​​ $$\begin{aligned} & \tan ^{-1}\left(\frac{1}{n^2+n+1}\right)+\tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right)+\ldots \ldots+ \\ & \quad=\tan ^{-1}\left(\frac{\frac{n+19}{n+21}-\frac{n-1}{n+1}}{1+\frac{n+19}{n+21} \times \frac{n-1}{n+1}}\right) \\ & \quad=\tan ^{-1} \frac{1}{1+(n+19)(n+20)} \frac{20}{n^2+20 n+1}=\mathrm{S} \\ & \therefore \quad \tan ^{-1} \mathrm{~S}=\frac{20}{n^2+20 n+1} \end{aligned}$$​tan−1(n2+n+11​)+tan−1(n2+3n+31​)+……+=tan−1(1+n+21n+19​×n+1n−1​n+21n+19​−n+1n−1​​)=tan−11+(n+19)(n+20)1​n2+20n+120​=S∴tan−1 S=n2+20n+120​​