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

<p>$$\begin{aligned} \mathrm{a}_{\mathrm{n}} & =\frac{\mathrm{n}^2+5 \mathrm{n}+6}{4} \\ \mathrm{~S}_{\mathrm{n}} & =\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{\mathrm{a}_{\mathrm{k}}}=\sum_1^{\mathrm{n}} \frac{4}{\mathrm{k}^2+5 \mathrm{k}+6} \\ & =4 \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{(\mathrm{k}+2)(\mathrm{k}+3)} \\ & =4 \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{\mathrm{k}+2}-\frac{1}{\mathrm{k}+3} \\ & =4\left(\frac{1}{3}-\frac{1}{4}\right)+4\left(\frac{1}{4}-\frac{1}{5}\right)+\ldots \ldots . . \end{aligned}$$</p> <p>$$\begin{array}{l} 4\left(\frac{1}{\mathrm{n}+2}-\frac{1}{\mathrm{n}+3}\right) \\ =4\left(\frac{1}{3}-\frac{1}{\mathrm{n}+3}\right) \\ =\frac{4 \mathrm{n}}{3(\mathrm{n}+3)} \\ { }^{507} \mathrm{~S}_{2025}=\frac{(507)(4)(2025)}{3(2028)} \\ =675 \end{array}$$</p>