For positive integers $n$, if $4 a_n=\left(n^2+5 n+6\right)$ and…

For positive integers $n$, if $4 a_n=\left(n^2+5 n+6\right)$ and $S_n=\sum_{k=1}^n\left(\frac{1}{a_k}\right)$, then the value of $507 S_{2025}$ is :
  1. $540$
  2. $675$
  3. $1350$
  4. $135$

Solution

$\begin{aligned} & S_n=\sum_{k=1}^n \frac{4}{K^2+5 k+6} \\ & =\sum_{k=1}^n \frac{4}{(K+2)(K+3)}=4 \sum_{K=1}^n\left(\frac{1}{K+2}-\frac{1}{K+3}\right)\end{aligned}$
$\begin{aligned} & =4\left[\frac{1}{3}-\frac{1}{4}\right] \\ & =4\left[\frac{1}{4}-\frac{1}{5}\right] \\ & =4\left[\frac{1}{n+2}-\frac{1}{n+3}\right] \\ & S_n=4\left[\frac{1}{3}-\frac{1}{n+3}\right]\end{aligned}$
$\begin{aligned} & S_{2025}=4\left[\frac{1}{3}-\frac{1}{2028}\right] \\ & S_{2025}=4\left[\frac{675}{2028}\right] \\ & 507 S_{2025}=675\end{aligned}$ *

Asked in: JEE Main 2025 (28 Jan Shift 2)

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