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📰 Solution: Assume $ h(x) = ax^2 + bx + c $. Using $ h(1) = a + b + c = 4 $, $ h(2) = 4a + 2b + c = 11 $, $ h(3) = 9a + 3b + c = 22 $. Subtract first equation from second: $ 3a + b = 7 $. Subtract second from third: $ 5a + b = 11 $. Subtract these: $ 2a = 4 \Rightarrow a = 2 $. Then $ 3(2) + b = 7 \Rightarrow b = 1 $. From $ 2 + 1 + c = 4 \Rightarrow c = 1 $. Thus, $ h(x) = 2x^2 + x + 1 $, and $ h(4) = 32 + 4 + 1 = 37 $. \boxed{37} 📰 Question: Compute $ \sum_{k=1}^{50} \frac{1}{k(k+2)} $. 📰 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}} 📰 Cabana Beach Apartment Gainesville Fl 393124 📰 Best Games Of 2025 Revealed If You Miss These Youll Be B 3156354 📰 Aja Volkman 7542585 📰 Jessica Sipos 2707804 📰 Dover Delaware 1039322 📰 Is This Oregon County Map The Key To Hidden Nature Trailcrets 71060 📰 Filter For Hard Water 2263013 📰 Gastric Bezoar 7537196 📰 Interest Rate On Money Market Account Bank Of America 480368 📰 China Harbor 9626542 📰 Geting Over It 6625248 📰 Crazy Car Games That Will Make You Scream With Extreme Fun 566639 📰 You Wont Believe How Easily You Can Convert Fahrenheit To Celsius 100800 📰 The Long Drive Game 1994695 📰 Demon Slayer Hashira Training Wallbreak Unbelievable Prep Resources Revealed 5667299