JEE Main 2025January, Unspecified (session-level dataset)MathematicsMediumSequence and Series179× since 2002If ∑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}∑r=1nTr=64(2n−1)(2n+1)(2n+3)(2n+5), then limn→∞∑r=1n(1Tr)\lim_{n \to \infty} \sum_{r=1}^n \left(\frac{1}{T_r}\right)limn→∞∑r=1n(Tr1) is equal to:AOption A. 0BOption B. 23\frac{2}{3}32COption C. 1DOption D. 13\frac{1}{3}31CheckStep-by-step hintSkip · show solution
