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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}} 📰 What Time Mcdonalds Breakfast End 4103948 📰 Unlock Efficient Accounting In Microsoft Officelife Saver For Excel Users 8924975 📰 Shinobi Demo 📰 Pagar Verizon Internet 📰 Pc And Playstation Crossplay Games 6416823 📰 Big Update Wells Fargo Calling Center And The Details Emerge 📰 Data Shows Marriott Points Value And The Story Spreads 📰 Usps Holidays 2025 9063600 📰 Labubu Cake Hack Swoon Sultry Sweetness Thats Going Viral Overnight 9095702 📰 Deltarune Walk Where The Stars Dont Shine 📰 Where Do I Find My Epic Id 9673276 📰 Hentai Loly 📰 Verizon Wireless Portable Charger 📰 Stick Project 📰 Pool Vacuums