Let be consecutive natural numbers. Then is equal to

Question

Let $a_{1} = 1, a_{2}, a_{3}, a_{4}, \dots$ be consecutive natural numbers. Then $\tan^{- 1} (\frac{1}{1 + a_{1} a_{2}}) + \tan^{- 1} (\frac{1}{1 + a_{2} a_{3}}) + \dots + \tan^{- 1} (\frac{1}{1 + a_{2024} a_{2025}})$ is equal to

Smart Achievers · Accepted Answer

(2)

By given condition, $a_{2} - a_{1} = a_{3} - a_{2} = \dots = a_{2025} - a_{2024} = 1$

$∴$ $\tan^{- 1} (\frac{a_{2} - a_{1}}{1 + a_{1} a_{2}}) + \tan^{- 1} (\frac{a_{3} - a_{2}}{1 + a_{2} a_{3}}) + \dots + \tan^{- 1} (\frac{a_{2025} - a_{2024}}{1 + a_{2024} a_{2025}})$

$[\tan^{- 1} a_{2} - \tan^{- 1} a_{1}] + [\tan^{- 1} a_{3} - \tan^{- 1} a_{2}] + \dots + [\tan^{- 1} a_{2025} - \tan^{- 1} a_{2024}]$

$= \tan^{- 1} a_{2025} - \tan^{- 1} a_{1} = \tan^{- 1} (2025) - \frac{π}{4}$

Solution

Q.
Let $a_{1} = 1, a_{2}, a_{3}, a_{4}, \dots$ be consecutive natural numbers. Then $\tan^{- 1} (\frac{1}{1 + a_{1} a_{2}}) + \tan^{- 1} (\frac{1}{1 + a_{2} a_{3}}) + \dots + \tan^{- 1} (\frac{1}{1 + a_{2024} a_{2025}})$ is equal to

1 $\frac{π}{4} + \cot^{- 1} (2025)$

2 $\tan^{- 1} (2025) - \frac{π}{4}$

3 $\tan^{- 1} (2024) - \frac{π}{4}$

4 $\frac{π}{4} - \tan^{- 1} (2024)$

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Solution

Q. Let a1=1,a2,a3,a4,… be consecutive natural numbers. Then tan-1(11+a1a2)+tan-1(11+a2a3)+…+tan-1(11+a2024a2025) is equal to

1 π4+cot-1(2025)

2 tan-1(2025)-π4

3 tan-1(2024)-π4

4 π4-tan-1(2024)