2025 AMC 10B Problem 15

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

Factor the denominator as

k(k + 2)(k + 4)

and use partial fractions

The pieces

\tfrac{\frac{1}{8}}{k} - \tfrac{\frac{1}{4}}{k + 2} + \tfrac{\frac{1}{8}}{k + 4}

telescope; only

k = 1, 2, 3, 4

leave a remainder

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.

Problem 15 in other years