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Sum of a telescoping series of inverse tangents (write each term as tan^-1 a - tan^-1 b)

This type of question, asked 10 times: 2021 Paper 1, 2021 Paper 5, 2021 Paper 7, 2021 Paper 19, 27 Jun 2022, 28 Jun 2022, 30 Jan 2023, 24 Jan 2025, 4 Apr 2025, 5 Apr 2026 S1. All 10 questions 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

    2021 Paper 5

    80
    Let Sk=∑r=1ktan⁡−1(6r22r+1+32r+1)S_{k}=\sum_{r = 1}^{k} \tan^{- 1}\left( \frac{6^{r}}{2^{2r + 1}+3^{2r + 1}} \right). Then lim⁡k→∞Sk\lim_{k \rightarrow \infty} S_{k} is equal to :

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  3. Question 3

    2021 Paper 7

    65
    If cot⁡−1(α)=cot⁡−12+cot⁡−18+cot⁡−118\cot^{- 1}(\alpha) =\cot^{- 1}2 +\cot^{- 1}8 +\cot^{- 1}18 +cot⁡−132+…+\cot^{- 1}32 + \ldots. upto 100 terms, then α\alpha is :

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  4. Question 4

    2021 Paper 19

    70
    If ∑r=150tan⁡−112r2=p\sum_{r = 1}^{50} \tan^{- 1}\frac{1}{2r^{2}}= p, then the value of tan⁡p\tan p is :

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  5. Question 5

    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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  6. Question 6

    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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  7. Question 7

    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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  8. Question 8

    24 Jan 2025

    128
    If α>β>γ>0\alpha > \beta > \gamma > 0, then the expression
    cot⁡−1{β+(1+β2)(α−β)}+cot⁡−1{γ+(1+γ2)(β−γ)}\cot^{- 1}\left\{ \beta +\frac{\left( 1 +\beta^{2} \right)}{(\alpha - \beta)} \right\}+\cot^{- 1}\left\{ \gamma +\frac{\left( 1 +\gamma^{2} \right)}{(\beta - \gamma)} \right\}
    +cot⁡−1{α+(1+α2)(γ−α)}+\cot^{- 1}\left\{ \alpha +\frac{\left( 1 +\alpha^{2} \right)}{(\gamma - \alpha)} \right\} is equal to:

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  9. Question 9

    4 Apr 2025

    128
    The sum of the infinite series cot⁡−1(74)+cot⁡−1(194)+cot⁡−1(394)+cot⁡−1(674)+…\cot^{- 1}\left( \frac{7}{4} \right)+\cot^{- 1}\left( \frac{19}{4} \right)+\cot^{- 1}\left( \frac{39}{4} \right)+\cot^{- 1}\left( \frac{67}{4} \right)+ \ldots. is :

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  10. Question 10

    5 Apr 2026 S1

    74
    If π4+∑p=111tan⁡−1(2p−11+22p−1)=α\frac{\pi}{4} + \sum_{p = 1}^{11} \tan^{- 1}\left( \frac{2^{p - 1}}{1 + 2^{2p - 1}} \right) = \alpha, then tan⁡α\tan\alpha is equal to ____\_\_\_\_ .