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

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(A) Given the expression: $\sum_{k=1}^n 	an^{-1} \left( \frac{1}{k^2+k+1} ight) = 	an^{-1} 	heta$. We know the identity $	an^{-1} x - 	an^{-1}

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$\frac{n}{n+2}$

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(A) Given the expression: $\sum_{k=1}^n 	an^{-1} \left( \frac{1}{k^2+k+1} ight) = 	an^{-1} 	heta$. We know 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 as: $\frac{1}{1+k(k+1)} = \frac{(k+1)-k}{1+k(k+1)}$. Thus,$	an^{-1} \left( \frac{1}{k^2+k+1} ight) = 	an^{-1} (k+1) - 	an^{-1} k$. Now,the summation becomes a telescoping series: $\sum_{k=1}^n (	an^{-1} (k+1) - 	an^{-1} k) = (	an^{-1} 2 - 	an^{-1} 1) + (	an^{-1} 3 - 	an^{-1} 2) + \dots + (	an^{-1} (n+1) - 	an^{-1} n)$. After canceling the intermediate terms,we are left with: $	an^{-1} (n+1) - 	an^{-1} 1 = 	an^{-1} 	heta$. Using the formula $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x-y}{1+xy} ight)$: $	an^{-1} \left( \frac{(n+1)-1}{1+(n+1)(1)} ight) = 	an^{-1} 	heta$. $	an^{-1} \left( \frac{n}{n+2} ight) = 	an^{-1} 	heta$. Therefore,$	heta = \frac{n}{n+2}$.

If $\sum_{k=1}^n \tan^{-1} \left( \frac{1}{k^2+k+1} \right) = \tan^{-1} ( \theta )$,then $\theta =$

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