JEE Mains · Maths · STD 11 - 8. sequence and series
If \(\sum_{r=1}^n T_r=\frac{(2 n-1)(2 n+1)(2 n+3)(2 n+5)}{64}\), then \(\lim _{n \rightarrow \infty} \sum_{r=1}^n\left(\frac{1}{T_r}\right)\) is equal to :
- A \(0\)
- B \(\frac{2}{3}\)
- C \(1\)
- D \(\frac{1}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{2}{3}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{T}_{\mathrm{n}}=\mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1} \\ & \Rightarrow \mathrm{~T}_{\mathrm{n}}=\frac{1}{8}(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3) \\ & \Rightarrow \frac{1}{\mathrm{~T}_{\mathrm{n}}}=\frac{8}{(2 \mathrm{n}-1)(2…
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