2025 AMC 10B Problem 15

Factor $k^3 + 6k^2 + 8k = k(k + 2)(k + 4),$ then split into partial fractions: $\dfrac{1}{k(k + 2)(k + 4)} = \dfrac{\frac{1}{8}}{k}$ $- \dfrac{\frac{1}{4}}{k + 2}$ $+ \dfrac{\frac{1}{8}}{k + 4}.$ Summing over all $k,$ the coefficient of $\tfrac1n$ cancels for $n \ge 5,$ so only the first few terms survive: $\tfrac18\left(1 + \tfrac12 - \tfrac13 - \tfrac14\right)$ $= \tfrac18 \cdot \tfrac{11}{12} = \tfrac{11}{96}.$ So $a + b = 11 + 96 = 107.$ Thus, D is the correct answer.