Question:

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

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When dealing with series sums, consider breaking the series into partial fractions to simplify the terms and cancel out intermediate terms.
Updated On: Feb 5, 2025
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The Correct Option is C

Solution and Explanation

We are given that: \[ a_n = \frac{n^2 + 5n + 6}{4} \] Now, we calculate \( S_n \) as: \[ S_n = \sum_{k=1}^{n} \frac{1}{a_k} = \sum_{k=1}^{n} \frac{4}{k^2 + 5k + 6} \] We can break this sum down into partial fractions: \[ S_n = 4 \sum_{k=1}^{n} \frac{1}{(k+2)(k+3)} \] This is equivalent to: \[ S_n = 4 \sum_{k=1}^{n} \left( \frac{1}{k+2} - \frac{1}{k+3} \right) \] Thus, we have: \[ S_n = 4 \left( \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \ldots \right) \] The sum for \( S_n \) will cancel out and simplify as follows: \[ S_{2025} = 4 \left( \frac{1}{3} - \frac{1}{2028} \right) \] Now calculate the value of \( 507 S_{2025} \): \[ 507 S_{2025} = 507 \times 4 \times \left( \frac{1}{3} - \frac{1}{2028} \right) = 675 \]
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