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\[ 250 + 30 = 280 \] - AMAZONAWS
\[ 250 + 30 = 280 \] - AMAZONAWS
📅 March 6, 2026
👤 scraface
Mar 06, 2026
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📌 The roots are $ x = 1 $ (with multiplicity 2) and $ x = -2 $. Therefore, there are **two distinct real roots**, but **three real roots counting multiplicity**. Since the question asks for the number of real roots (not distinct), the answer is:
📌 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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![\[ 250 + 30 = 280 \]](https://soloferat.biz.id/images/-250--30--280-.jpg)
