If f(n) = tan left[tan ^{-1} frac{1}{1+2} + t | vedclass

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(C) Given $f(n) = 	an \left[\sum_{k=1}^{n} 	an ^{-1} \frac{1}{1+k(k+1)}ight]$.We know that $	an ^{-1} x - 	an ^{-1} y = 	an ^{-1} \left(\frac{x

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$\frac{2021}{2023}$

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(C) Given $f(n) = 	an \left[\sum_{k=1}^{n} 	an ^{-1} \frac{1}{1+k(k+1)}ight]$.We know that $	an ^{-1} x - 	an ^{-1} y = 	an ^{-1} \left(\frac{x-y}{1+xy}ight)$.We can write the general term as $	an ^{-1} \left(\frac{(k+1)-k}{1+k(k+1)}ight) = 	an ^{-1}(k+1) - 	an ^{-1}(k)$.Thus,the sum becomes a telescoping series:$S_n = (	an ^{-1}(2) - 	an ^{-1}(1)) + (	an ^{-1}(3) - 	an ^{-1}(2)) + \ldots + (	an ^{-1}(n+1) - 	an ^{-1}(n))$.All intermediate terms cancel out,leaving $S_n = 	an ^{-1}(n+1) - 	an ^{-1}(1)$.Using the formula $	an ^{-1} A - 	an ^{-1} B = 	an ^{-1} \left(\frac{A-B}{1+AB}ight)$:$S_n = 	an ^{-1} \left(\frac{(n+1)-1}{1+(n+1)(1)}ight) = 	an ^{-1} \left(\frac{n}{n+2}ight)$.Therefore,$f(n) = 	an \left[	an ^{-1} \left(\frac{n}{n+2}ight)ight] = \frac{n}{n+2}$.For $n = 2021$,$f(2021) = \frac{2021}{2021+2} = \frac{2021}{2023}$.

If $f(n) = \tan \left[\tan ^{-1} \frac{1}{1+2} + \tan ^{-1} \frac{1}{1+6} + \tan ^{-1} \frac{1}{1+12} + \ldots + \tan ^{-1} \frac{1}{1+n(n+1)}\right]$,then $f(2021) =$

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