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Jee Main 2025 — Medium JEE math MCQ

If \(\sum_{r=1}^{n} T_r = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}\), then \(\lim_{n \to \infty} \sum_{r=1}^{n} \left( \frac{1}{T_r} \right)\) is equal to
  1. A. \(0\)
  2. B. \(\frac{4}{3}\)
  3. C. \(1\)
  4. D. \(\frac{1}{2}\)

Solution

The correct option is **B**. (B. \(\frac{4}{3}\))

MATH

medium • PYQ Reworded
Question
Read carefully, then pick the best option.
If ∑r=1nTr=(2n−1)(2n+1)(2n+3)(2n+5)64\sum_{r=1}^{n} T_r = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}, then lim⁡n→∞∑r=1n(1Tr)\lim_{n \to \infty} \sum_{r=1}^{n} \left( \frac{1}{T_r} \right) is equal to
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Jee Main 2025 — Medium JEE Mathematics MCQ | MyGoalPrep