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JEE Main 202528 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Sequences and Series

For positive integers n , if 4 a_n=(n^2+5 n+6) and S_n=∑_ k=1...

For positive integers nn, if 4an=(n2+5n+6)4 a_n=\left(n^2+5 n+6\right) and S_n=\sum_{k=1}^n\left(\frac{1}{a_k}\right),thenthevalueof, then the value of507 S_{2025} is :

Options

  1. A540540
  2. B675675
  3. C13501350
  4. D135135

Correct answer

B. 675675

Step-by-step solution

Sn=∑k=1n4K2+5k+6=∑k=1n4(K+2)(K+3)=4∑K=1n(1K+2−1K+3)\begin{aligned} & S_n=\sum_{k=1}^n \frac{4}{K^2+5 k+6} \\ & =\sum_{k=1}^n \frac{4}{(K+2)(K+3)}=4 \sum_{K=1}^n\left(\frac{1}{K+2}-\frac{1}{K+3}\right)\end{aligned}=4[13−14]=4[14−15]=4[1n+2−1n+3]Sn=4[13−1n+3]\begin{aligned} & =4\left[\frac{1}{3}-\frac{1}{4}\right] \\ & =4\left[\frac{1}{4}-\frac{1}{5}\right] \\ & =4\left[\frac{1}{n+2}-\frac{1}{n+3}\right] \\ & S_n=4\left[\frac{1}{3}-\frac{1}{n+3}\right]\end{aligned} S2025=4[13−12028]S2025=4[6752028]507S2025=675\begin{aligned} & S_{2025}=4\left[\frac{1}{3}-\frac{1}{2028}\right] \\ & S_{2025}=4\left[\frac{675}{2028}\right] \\ & 507 S_{2025}=675\end{aligned}$
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