The sum to infinity of cot⁻¹(2k²) for k=1, 2, 3... is:
The sum to infinity of cot⁻¹(2k²) for k=1, 2, 3... is:
- A. π/4
- B. π/2
- C. π
- D. 3π/4
Answer: A) π/4
Explanation: cot⁻¹(2k²) = tan⁻¹(1 / 2k²) = tan⁻¹(2 / 4k²) = tan⁻¹( (2k+1 − (2k−1)) / (1 + (2k+1)(2k−1)) ) = tan⁻¹(2k+1) − tan⁻¹(2k−1). Sum = tan⁻¹(∞) − tan⁻¹(1) = π/4.
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