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Telescoping sum of tan^-1(1/(1 + r + r^2)) (finite or infinite)

The same question, asked 4 times: 2021 Paper 1*, 27 Jun 2022*, 28 Jun 2022*, 30 Jan 2023. Every wording below, oldest first.

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  1. Question 1

    2021 Paper 1

    85
    lim⁡n→∞tan⁡{∑r=1ntan⁡−1(11+r+r2)}\lim_{n \rightarrow \infty} \tan\left\{ \sum_{r= 1}^{n} \tan^{- 1}\left( \frac{1}{1 +r+r^{2}} \right) \right\} is equal to

  2. Question 2

    27 Jun 2022

    168
    The value of cot⁡(∑n=150tan⁡−1(11+n+n2))\cot\left( \sum_{n = 1}^{50} \tan^{- 1}\left( \frac{1}{1 + n +n^{2}} \right) \right) is

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  3. Question 3

    28 Jun 2022

    168
    The value of lim⁡n→∞6tan⁡{∑r=1ntan⁡−1(1r2+3r+3)}\lim_{n \rightarrow \infty} 6\tan\left\{ \sum_{r = 1}^{n} \tan^{- 1}\left( \frac{1}{r^{2}+ 3r + 3} \right) \right\} is equal to

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  4. Question 4

    30 Jan 2023

    153
    Let a1=1,a2,a3,a4,….a_{1}= 1,a_{2},a_{3},a_{4},\ldots.. be consecutive natural numbers. Then tan⁡−1(11+a1a2)+tan⁡−1(11+a2a3)+⋯⋅+tan⁡−1(11+a2021a2022)\tan^{- 1}\left( \frac{1}{1 +a_{1}a_{2}} \right)+\tan^{- 1}\left( \frac{1}{1 +a_{2}a_{3}} \right)+ \cdots \cdot +\tan^{- 1}\left( \frac{1}{1 +a_{2021}a_{2022}} \right) is equal to

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