If sum_{n=1}^{2026} tan^{-1}(frac{1}{n^2+n+1} | vedclass

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(C) We use the identity $	an^{-1}(a) - 	an^{-1}(b) = 	an^{-1}(\frac{a-b}{1+ab})$.Note that $\frac{1}{n^2+n+1} = \frac{(n+1)-n}{1+n(n+1)}$.Thus, $

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$1014$

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(C) We use the identity $	an^{-1}(a) - 	an^{-1}(b) = 	an^{-1}(\frac{a-b}{1+ab})$.Note that $\frac{1}{n^2+n+1} = \frac{(n+1)-n}{1+n(n+1)}$.Thus, $	an^{-1}(\frac{1}{n^2+n+1}) = 	an^{-1}(n+1) - 	an^{-1}(n)$.The sum is $\sum_{n=1}^{2026} (	an^{-1}(n+1) - 	an^{-1}(n))$.This is a telescoping sum: $(	an^{-1}(2) - 	an^{-1}(1)) + (	an^{-1}(3) - 	an^{-1}(2)) + \dots + (	an^{-1}(2027) - 	an^{-1}(2026))$.The sum simplifies to $	an^{-1}(2027) - 	an^{-1}(1)$.Using the identity again: $	an^{-1}(2027) - 	an^{-1}(1) = 	an^{-1}(\frac{2027-1}{1+2027 \cdot 1}) = 	an^{-1}(\frac{2026}{2028}) = 	an^{-1}(\frac{1013}{1014})$.We are given $	an^{-1}(1 - \frac{1}{x}) = 	an^{-1}(\frac{1013}{1014})$.So, $1 - \frac{1}{x} = \frac{1013}{1014} \implies \frac{1}{x} = 1 - \frac{1013}{1014} = \frac{1}{1014}$.Therefore, $x = 1014$.

If $\sum_{n=1}^{2026} \tan^{-1}(\frac{1}{n^2+n+1}) = \tan^{-1}(1 - \frac{1}{x})$, where $x \neq 0$, then $x = $

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