Let a_1=1, a_2, a_3, a_4, ldots be consecutiv | vedclass

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(C) Given that $a_1, a_2, \ldots, a_{2022}$ are consecutive natural numbers,we have $a_{k+1} - a_k = 1$ for all $k \ge 1$.Since $a_1 = 1$,we have $a_2

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$	an ^{-1}(2022)-\frac{\pi}{4}$

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(C) Given that $a_1, a_2, \ldots, a_{2022}$ are consecutive natural numbers,we have $a_{k+1} - a_k = 1$ for all $k \ge 1$.Since $a_1 = 1$,we have $a_2 = 2, a_3 = 3, \ldots, a_{2022} = 2022$.The general term of the series is $	an ^{-1}\left(\frac{1}{1+ a _k a _{k+1}}ight)$.Since $a_{k+1} - a_k = 1$,we can write the term as $	an ^{-1}\left(\frac{a_{k+1} - a_k}{1+ a _k a _{k+1}}ight)$.Using the identity $	an ^{-1} x - 	an ^{-1} y = 	an ^{-1}\left(\frac{x-y}{1+xy}ight)$,the series becomes:$\sum_{k=1}^{2021} (	an ^{-1} a_{k+1} - 	an ^{-1} a_k) = (	an ^{-1} a_2 - 	an ^{-1} a_1) + (	an ^{-1} a_3 - 	an ^{-1} a_2) + \ldots + (	an ^{-1} a_{2022} - 	an ^{-1} a_{2021})$.This is a telescoping series,which simplifies to $	an ^{-1} a_{2022} - 	an ^{-1} a_1$.Substituting the values $a_{2022} = 2022$ and $a_1 = 1$,we get $	an ^{-1}(2022) - 	an ^{-1}(1) = 	an ^{-1}(2022) - \frac{\pi}{4}$.

Let $a_1=1, a_2, a_3, a_4, \ldots$ be consecutive natural numbers. Then $\tan ^{-1}\left(\frac{1}{1+ a _1 a _2}\right)+\tan ^{-1}\left(\frac{1}{1+ a _2 a _3}\right)+\ldots+\tan ^{-1}\left(\frac{1}{1+ a _{2021} a _{2022}}\right)$ is equal to

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