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If $\sum_{n=1}^k \tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right)=\tan ^{-1} \alpha$, then $\alpha=$
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Verified Answer
The correct answer is:
$\frac{k}{2 k+5}$
We have,
$\begin{aligned}
\sum_{n=1}^k \tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right) =\tan ^{-1} \alpha \\
\sum_{n=1}^k \tan ^{-1}\left(\frac{(n+2)-(n+1)}{1+(n+2)(n+1)}\right) =\tan ^{-1} \alpha \\
\Rightarrow \quad \sum_{n=1}^k\left(\tan ^{-1}(n+2)-\tan ^{-1}(n+1)\right) =\tan ^{-1} \alpha \\
\Rightarrow \quad\left(\tan ^{-1} 3-\tan ^{-1} 2\right)+\left(\tan ^{-1} 4-\tan ^{-1} 2\right) \ldots & \tan ^{-1}(k+2)-\tan ^{-1}(k+1) \\
\tan ^{-1}(k+2)-\tan ^{-1} 2 =\tan ^{-1} \alpha \\
\tan ^{-1}\left(\frac{k+2-2}{1+(k+2)(2)}\right) =\tan ^{-1} \alpha \\
\frac{k}{1+2 k+4}=\alpha \Rightarrow \alpha =\frac{k}{2 k+5}
\end{aligned}$
$\begin{aligned}
\sum_{n=1}^k \tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right) =\tan ^{-1} \alpha \\
\sum_{n=1}^k \tan ^{-1}\left(\frac{(n+2)-(n+1)}{1+(n+2)(n+1)}\right) =\tan ^{-1} \alpha \\
\Rightarrow \quad \sum_{n=1}^k\left(\tan ^{-1}(n+2)-\tan ^{-1}(n+1)\right) =\tan ^{-1} \alpha \\
\Rightarrow \quad\left(\tan ^{-1} 3-\tan ^{-1} 2\right)+\left(\tan ^{-1} 4-\tan ^{-1} 2\right) \ldots & \tan ^{-1}(k+2)-\tan ^{-1}(k+1) \\
\tan ^{-1}(k+2)-\tan ^{-1} 2 =\tan ^{-1} \alpha \\
\tan ^{-1}\left(\frac{k+2-2}{1+(k+2)(2)}\right) =\tan ^{-1} \alpha \\
\frac{k}{1+2 k+4}=\alpha \Rightarrow \alpha =\frac{k}{2 k+5}
\end{aligned}$
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