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Home » Why Percebes Are Worth Thousands: This Beach Staple Is Becoming Wildly Rare! - AMAZONAWS

Why Percebes Are Worth Thousands: This Beach Staple Is Becoming Wildly Rare! - AMAZONAWS

Why Percebes Are Worth Thousands: This Beach Staple Is Becoming Wildly Rare

📅 March 6, 2026 👤 scraface
Mar 06, 2026
Why Percebes Are Worth Thousands: This Beach Staple Is Becoming Wildly Rare!

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📌 P'(t) = 3t^2 - 12t + 9.
📌 Solution: Use partial fractions: $ \frac{1}{k(k+2)} = \frac{1}{2}\left( \frac{1}{k} - \frac{1}{k+2} \right) $. The sum becomes $ \frac{1}{2} \left( \sum_{k=1}^{50} \frac{1}{k} - \sum_{k=1}^{50} \frac{1}{k+2} \right) = \frac{1}{2} \left( \sum_{k=1}^{50} \frac{1}{k} - \sum_{k=3}^{52} \frac{1}{k} \right) $. Telescoping gives $ \frac{1}{2} \left( \frac{1}{1} + \frac{1}{2} - \frac{1}{51} - \frac{1}{52} \right) = \frac{1}{2} \left( \frac{3}{2} - \frac{103}{2652} \right) = \frac{1}{2} \cdot \frac{3975}{2652} = \frac{1325}{1768} $. \boxed{\dfrac{1325}{1768}}
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