Verified Solution • 2025–26 Session moderate

QJEEM25Ma544b: JEE Main 2025 Mathematics - Class 12 JEEM25Ma544b Solution: If , then is equal to:

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:

Multiple Choice Options

  • A 13\frac{1}{3}
  • B 1
  • C 0
  • D 23\frac{2}{3}

Step-by-Step Verified Solution

amp;Tn=Sn−Sn−1amp;⇒Tn=18(2n−1)(2n+1)(2n+3)amp;⇒1Tn=8(2n−1)(2n+1)(2n+3)amp;lim⁡n→∞∑r=1n1Tr=lim⁡n→∞8∑r=1n1(2n−1)(2n+1)(2n+3)amp;=lim⁡n→∞84∑(1(2n−1)(2n+1)−1(2n+1)(2n+3))amp;=lim⁡n→∞2[(11.3−13.5)+(13.5−15.7)+…]amp;=23\begin{aligned} & T_n=S_n-S_{n-1} \\\\ & \Rightarrow T_n=\frac{1}{8}(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3) \\\\ & \Rightarrow \frac{1}{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}{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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