Key Concepts and Formulas

• Quadratic Factorization: Factoring a quadratic expression $ax^2+bx+c$ into its linear factors.
• Partial Fraction Decomposition: Expressing a rational function as a sum of simpler rational functions.
• Telescoping Series: A series where most terms cancel out, simplifying the sum to a few initial and final terms.

Step-by-Step Solution

1. Express $a_n$ explicitly: We are given the relation $4 a_n = n^2 + 5n + 6$. To find $a_n$, we divide by 4: $a_n = \frac{n^2 + 5n + 6}{4}$ This step isolates the term $a_n$, which is necessary for subsequent calculations.

2. Express $\frac{1}{a_k}$: The sum $S_n$ is defined as $S_n = \sum_{k=1}^n \frac{1}{a_k}$. We substitute the expression for $a_k$: $S_n = \sum_{k=1}^n \frac{1}{\frac{k^2 + 5k + 6}{4}} = \sum_{k=1}^n \frac{4}{k^2 + 5k + 6}$ This step rewrites the summand in terms of $k$, preparing it for simplification.

3. Factor the quadratic expression: The denominator $k^2 + 5k + 6$ is a quadratic expression. We factor it to identify potential terms for partial fraction decomposition: $k^2 + 5k + 6 = (k+2)(k+3)$ So, the expression for $S_n$ becomes: $S_n = \sum_{k=1}^n \frac{4}{(k+2)(k+3)}$ Factoring the denominator is crucial for applying partial fraction decomposition.

4. Apply Partial Fraction Decomposition: We decompose the fraction $\frac{1}{(k+2)(k+3)}$ into simpler terms: $\frac{1}{(k+2)(k+3)} = \frac{A}{k+2} + \frac{B}{k+3}$ Multiplying both sides by $(k+2)(k+3)$ gives: $1 = A(k+3) + B(k+2)$ To find $A$, we set $k = -2$: $1 = A(-2+3) + B(-2+2) \Rightarrow 1 = A$. To find $B$, we set $k = -3$: $1 = A(-3+3) + B(-3+2) \Rightarrow 1 = -B \Rightarrow B = -1$. Thus, the decomposition is: $\frac{1}{(k+2)(k+3)} = \frac{1}{k+2} - \frac{1}{k+3}$ This decomposition allows us to express the summand as a difference of two terms, which is essential for a telescoping series.

5. Rewrite the Sum using Partial Fractions: Substitute the partial fraction decomposition back into the expression for $S_n$: $S_n = 4 \sum_{k=1}^n \left(\frac{1}{k+2} - \frac{1}{k+3}\right)$ This step prepares the sum for the telescoping cancellation.

6. Expand the Sum (Telescoping Series): We expand the sum to visualize the cancellation of terms: $S_n = 4 \left[ \left(\frac{1}{1+2} - \frac{1}{1+3}\right) + \left(\frac{1}{2+2} - \frac{1}{2+3}\right) + \left(\frac{1}{3+2} - \frac{1}{3+3}\right) + \cdots + \left(\frac{1}{n+2} - \frac{1}{n+3}\right) \right]$ $S_n = 4 \left[ \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{6}\right) + \cdots + \left(\frac{1}{n+2} - \frac{1}{n+3}\right) \right]$ The middle terms cancel out in pairs.

7. Simplify the Telescoping Series: After cancellation, only the first part of the first term and the second part of the last term remain: $S_n = 4 \left(\frac{1}{3} - \frac{1}{n+3}\right)$ To simplify further, we find a common denominator: $S_n = 4 \left(\frac{(n+3) - 3}{3(n+3)}\right) = 4 \left(\frac{n}{3(n+3)}\right) = \frac{4n}{3(n+3)}$ This provides a compact formula for $S_n$.

8. Calculate $S_{2025}$: We need to find the value of $S_n$ when $n = 2025$. Substitute $n=2025$ into the formula for $S_n$: $S_{2025} = \frac{4 \cdot 2025}{3(2025+3)} = \frac{4 \cdot 2025}{3 \cdot 2028}$ This step calculates the specific sum required.

9. Calculate $507 S_{2025}$: Finally, we need to compute $507 S_{2025}$: $507 S_{2025} = 507 \cdot \frac{4 \cdot 2025}{3 \cdot 2028}$ We can simplify this expression. Notice that $2028 = 4 \times 507$. Substituting this into the equation: $507 S_{2025} = 507 \cdot \frac{4 \cdot 2025}{3 \cdot (4 \cdot 507)}$ Cancel out common factors: $507 S_{2025} = \frac{2025}{3}$ Performing the division: $507 S_{2025} = 675$ This step yields the final numerical value.

Common Mistakes & Tips

• Forgetting the Constant Factor: Ensure the factor of 4 is carried through the partial fraction decomposition and the sum.
• Incorrect Cancellation in Telescoping Series: Carefully write out the first few and last few terms of the series to confirm which terms cancel and which remain.
• Arithmetic Errors: Double-check all arithmetic calculations, especially when simplifying fractions and performing the final multiplication.

Summary

The problem requires finding the value of $507 S_{2025}$, where $S_n$ is a sum involving the reciprocal of $a_n$. We first expressed $a_n$ and then $\frac{1}{a_k}$. By factoring the denominator of $\frac{1}{a_k}$ and applying partial fraction decomposition, we transformed the sum into a telescoping series. Evaluating the telescoping series yielded a simplified expression for $S_n$. Substituting $n=2025$ and multiplying by 507 gave the final answer.

The final answer is \boxed{675}.