We are given that: \[ a_n = \frac{n^2 + 5n + 6}{4} \] To calculate \( S_n \): \[ S_n = \sum_{k=1}^{n} \frac{1}{a_k} = \sum_{k=1}^{n} \frac{4}{k^2 + 5k + 6} \]
Step 2: Break the Sum into Partial FractionsDecomposing the given expression into partial fractions: \[ S_n = 4 \sum_{k=1}^{n} \frac{1}{(k+2)(k+3)} \] Which simplifies to: \[ S_n = 4 \sum_{k=1}^{n} \left( \frac{1}{k+2} - \frac{1}{k+3} \right) \]
Step 3: Evaluate the SeriesThe series telescopes, giving: \[ S_n = 4 \left( \frac{1}{3} - \frac{1}{n+3} \right) \] For \( n = 2025 \), \[ S_{2025} = 4 \left( \frac{1}{3} - \frac{1}{2028} \right) \]
Step 4: Compute \( 507 \times S_{2025} \)Multiplying by 507: \[ 507 S_{2025} = 507 \times 4 \times \left( \frac{1}{3} - \frac{1}{2028} \right) \] Simplifying, \[ 507 S_{2025} = 675 \]
Final Answer: 675If \(\sum\)\(_{r=1}^n T_r\) = \(\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}\) , then \( \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{T_r} \) is equal to :