QJEEM25Mae7a2: JEE Main 2025 Mathematics - Class 11 JEEM25Mae7a2 Solution: For positive integers , if and , then the value of is :

$\begin{aligned}\mathrm{a}_{\mathrm{n}} & =\frac{\mathrm{n}^2+5 \mathrm{n}+6}{4} \\\mathrm{~S}_{\mathrm{n}} & =\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{\mathrm{a}_{\mathrm{k}}}=\sum_1^{\mathrm{n}} \frac{4}{\mathrm{k}^2+5 \mathrm{k}+6} \\& =4 \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{(\mathrm{k}+2)(\mathrm{k}+3)} \\& =4 \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{\mathrm{k}+2}-\frac{1}{\mathrm{k}+3} \\& =4\left(\frac{1}{3}-\frac{1}{4}\right)+4\left(\frac{1}{4}-\frac{1}{5}\right)+\ldots \ldots . .\end{aligned}$

$\begin{array}{l}4\left(\frac{1}{\mathrm{n}+2}-\frac{1}{\mathrm{n}+3}\right) \\=4\left(\frac{1}{3}-\frac{1}{\mathrm{n}+3}\right) \\=\frac{4 \mathrm{n}}{3(\mathrm{n}+3)} \\{ }^{507} \mathrm{~S}_{2025}=\frac{(507)(4)(2025)}{3(2028)} \\=675\end{array}$