jee-main 2023 Q76

jee-main · India · session1_30jan_shift2 Arithmetic Sequences and Series Telescoping or Non-Standard Summation Involving an AP

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
(1) $\frac{\pi}{4} - \cot^{-1}(2022)$
(2) $\cot^{-1}(2022) - \frac{\pi}{4}$
(3) $\tan^{-1}(2022) - \frac{\pi}{4}$
(4) $\frac{\pi}{4} - \tan^{-1}(2022)$

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\\
(1) $\frac{\pi}{4} - \cot^{-1}(2022)$\\
(2) $\cot^{-1}(2022) - \frac{\pi}{4}$\\
(3) $\tan^{-1}(2022) - \frac{\pi}{4}$\\
(4) $\frac{\pi}{4} - \tan^{-1}(2022)$

Paper Questions

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