Question

The sum

$\sum_{k=1}^{\infty} \frac{1}{k^3 + 6k^2 + 8k}$

can be expressed as $\dfrac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. What is $a + b?$

$89$

$97$

$102$

$107$

$129$

Answer: D

Concepts:partial fractions telescoping

Difficulty rating: 1600

Solution:

Factor $k^3 + 6k^2 + 8k = k(k + 2)(k + 4),$ then split into partial fractions: $\dfrac{1}{k(k + 2)(k + 4)} = \dfrac{1/8}{k}$ $- \dfrac{1/4}{k + 2}$ $+ \dfrac{1/8}{k + 4}.$ 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.

2025 AMC 10B Problem 15

Answer: D

Difficulty rating: 1600

Solution:

Problem 15 in Other Years