JEE Main 2025 — 28 January, Evening Shift — Question 13

$a_{n}=\frac{n^{2}+5 n+6}{4}$

$$\begin{aligned} \mathrm{S}_{\mathrm{n}} & =\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{\mathrm{a}_{\mathrm{k}}}=\sum_{1}^{\mathrm{n}} \frac{4}{\mathrm{k}^{2}+5 \mathrm{k}+6} \\& =4 \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{(\mathrm{k}+2)(\mathrm{k}+3)} \\& =4 \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{\mathrm{k}+2}-\frac{1}{\mathrm{k}+3} \\& =4\left(\frac{1}{3}-\frac{1}{4}\right)+4\left(\frac{1}{4}-\frac{1}{5}\right)+\ldots \ldots . .\end{aligned}$$

$\begin{array}{r} 4\left(\frac{1}{n+2}-\frac{1}{n+3}\right) =4\left(\frac{1}{3}-\frac{1}{n+3}\right) =\frac{4 n}{3(n+3)}\\ { }{507} S_{2025}=\frac{(507)(4)(2025)}{3(2028)} =675 \end{array}$.