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: 675What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)