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$S=\tan ^{-1}\left(\frac{1}{n^2+n+1}\right)+\tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right)+\ldots$ $+\tan ^{-1}\left(\frac{1}{1+(n+19)(n+20)}\right)$, then $\tan S$ is equal to :
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Verified Answer
The correct answer is:
$\frac{20}{n^2+20 n+1}$
$\frac{20}{n^2+20 n+1}$
We know that,
$$
\begin{array}{r}
\tan ^{-1} \frac{1}{1+2}+\tan ^{-1} \frac{1}{1+2 \times 3}+\tan ^{-1} \frac{1}{1+3 \times 4}+\ldots \ldots+ \\
\tan ^{-1} \frac{1}{1+(n-1) n}+\tan ^{-1} \frac{1}{1+n(n+1)}+\ldots \ldots+ \\
\tan ^{-1} \frac{1}{1+(n+19)(n+20)}=\tan ^{-1} \frac{n+19}{n+21} \\
\Rightarrow \tan ^{-1} \frac{n-1}{n+1}+\tan ^{-1} \frac{1}{1+n(n+1)}+\tan ^{-1} \frac{1}{1+(n+1)(n+2)} \\
+\ldots \ldots+\frac{1}{1+(n+19)(n+20)}=\tan ^{-1} \frac{n+19}{n+21} \\
\tan ^{-1} \frac{1}{1+n(n+1)}+\tan ^{-1} \frac{1}{1+(n+1)(n+2)}+\ldots \ldots+ \\
\frac{1}{1+(n+19)(n+20)}=\tan ^{-1} \frac{n+19}{n+21}-\tan ^{-1} \frac{n-1}{n+1}
\end{array}
$$
$$
\begin{aligned}
& \tan ^{-1}\left(\frac{1}{n^2+n+1}\right)+\tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right)+\ldots \ldots+ \\
& \quad=\tan ^{-1}\left(\frac{\frac{n+19}{n+21}-\frac{n-1}{n+1}}{1+\frac{n+19}{n+21} \times \frac{n-1}{n+1}}\right) \\
& \quad=\tan ^{-1} \frac{1}{1+(n+19)(n+20)} \frac{20}{n^2+20 n+1}=\mathrm{S} \\
& \therefore \quad \tan ^{-1} \mathrm{~S}=\frac{20}{n^2+20 n+1}
\end{aligned}
$$
$$
\begin{array}{r}
\tan ^{-1} \frac{1}{1+2}+\tan ^{-1} \frac{1}{1+2 \times 3}+\tan ^{-1} \frac{1}{1+3 \times 4}+\ldots \ldots+ \\
\tan ^{-1} \frac{1}{1+(n-1) n}+\tan ^{-1} \frac{1}{1+n(n+1)}+\ldots \ldots+ \\
\tan ^{-1} \frac{1}{1+(n+19)(n+20)}=\tan ^{-1} \frac{n+19}{n+21} \\
\Rightarrow \tan ^{-1} \frac{n-1}{n+1}+\tan ^{-1} \frac{1}{1+n(n+1)}+\tan ^{-1} \frac{1}{1+(n+1)(n+2)} \\
+\ldots \ldots+\frac{1}{1+(n+19)(n+20)}=\tan ^{-1} \frac{n+19}{n+21} \\
\tan ^{-1} \frac{1}{1+n(n+1)}+\tan ^{-1} \frac{1}{1+(n+1)(n+2)}+\ldots \ldots+ \\
\frac{1}{1+(n+19)(n+20)}=\tan ^{-1} \frac{n+19}{n+21}-\tan ^{-1} \frac{n-1}{n+1}
\end{array}
$$
$$
\begin{aligned}
& \tan ^{-1}\left(\frac{1}{n^2+n+1}\right)+\tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right)+\ldots \ldots+ \\
& \quad=\tan ^{-1}\left(\frac{\frac{n+19}{n+21}-\frac{n-1}{n+1}}{1+\frac{n+19}{n+21} \times \frac{n-1}{n+1}}\right) \\
& \quad=\tan ^{-1} \frac{1}{1+(n+19)(n+20)} \frac{20}{n^2+20 n+1}=\mathrm{S} \\
& \therefore \quad \tan ^{-1} \mathrm{~S}=\frac{20}{n^2+20 n+1}
\end{aligned}
$$
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