If sum_{n=1}^k tan ^{-1}left(frac{1}{n^2+3 n+ | vedclass

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(C) We have, $\sum_{n=1}^k 	an ^{-1}\left(\frac{1}{n^2+3 n+3}ight) = 	an ^{-1} \alpha$. Using the identity $	an ^{-1} x - 	an ^{-1} y = 	an ^{-

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$\frac{k}{2 k+5}$

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(C) We have, $\sum_{n=1}^k 	an ^{-1}\left(\frac{1}{n^2+3 n+3}ight) = 	an ^{-1} \alpha$. Using the identity $	an ^{-1} x - 	an ^{-1} y = 	an ^{-1}\left(\frac{x-y}{1+xy}ight)$, we can rewrite the term inside the summation: $\frac{1}{n^2+3n+3} = \frac{(n+2)-(n+1)}{1+(n+2)(n+1)}$. Thus, the summation becomes: $\sum_{n=1}^k (	an ^{-1}(n+2) - 	an ^{-1}(n+1)) = 	an ^{-1} \alpha$. Expanding the sum: $(	an ^{-1} 3 - 	an ^{-1} 2) + (	an ^{-1} 4 - 	an ^{-1} 3) + \dots + (	an ^{-1}(k+2) - 	an ^{-1}(k+1)) = 	an ^{-1} \alpha$. This is a telescoping series, so all intermediate terms cancel out: $	an ^{-1}(k+2) - 	an ^{-1} 2 = 	an ^{-1} \alpha$. Applying the formula again: $	an ^{-1}\left(\frac{(k+2)-2}{1+(k+2)(2)}ight) = 	an ^{-1} \alpha$. $	an ^{-1}\left(\frac{k}{1+2k+4}ight) = 	an ^{-1} \alpha$. Therefore, $\alpha = \frac{k}{2k+5}$.

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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