If ∑_ r=1 ^n T_r=((2 n-1)(2 n+1)(2 n+3)(2 n+5))/(64) , then _ n arrow ∞ ∑_ r=1 ^n((1)/(T_r)) is equal to :

Correct answer: B. (2)/(3) aligned & T_ n =S_ n -S_ n-1 & ~T_ n =(1)/(8)(2 n-1)(2 n+1)(2 n+3) & 1 ~T_ n = 8 (2 n-1)(2 n+1)(2 n+3) & _ n arrow ∞ ∑_ r=1 ^ n 1 ~T_ r = _ n arrow ∞ 8 ∑_ r=1 ^ n 1 (2 n-1)(2 n+1)(2 n+3) & = _ n arrow ∞ (8)/(4) ∑( 1 (2 n-1)(2 n+1) - 1 (2 n+1)(2 n+3) ) & = _ n arrow ∞ 2[((1)/(1.3)-(1)/(3.5))+((1)/(3.5)-(1)/(5.7))+ ] & =(2)/(3) aligned

JEE Main 202522 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 22 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2025 · 22 January · Morning Shift · Mathematics · Sequences and Series

If ∑_ r=1 ^n T_r=((2 n-1)(2 n+1)(2 n+3)(2 n+5))/(64) , then _ n...

\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 :

Options

A $0$
B $\frac{2}{3}$
C $1$
D $\frac{1}{3}$

Correct answer

B. $\frac{2}{3}$

Step-by-step solution

\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 \mathrm{n}+1)(2 \mathrm{n}+3)} \\ & \lim _{\mathrm{n} \rightarrow \infty} \sum_{\mathrm{r}=1}^{\mathrm{n}} \frac{1}{\mathrm{~T}_{\mathrm{r}}}=\lim _{\mathrm{n} \rightarrow \infty} 8 \sum_{\mathrm{r}=1}^{\mathrm{n}} \frac{1}{(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3)} \\ & =\lim _{\mathrm{n} \rightarrow \infty} \frac{8}{4} \sum\left(\frac{1}{(2 \mathrm{n}-1)(2 \mathrm{n}+1)}-\frac{1}{(2 \mathrm{n}+1)(2 \mathrm{n}+3)}\right) \\ & =\lim _{\mathrm{n} \rightarrow \infty} 2\left[\left(\frac{1}{1.3}-\frac{1}{3.5}\right)+\left(\frac{1}{3.5}-\frac{1}{5.7}\right)+\ldots\right] \\ & =\frac{2}{3}\end{aligned}

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