AMC 10 · 2025 · #15

Grade 6 algebra

Archetype Telescoping Collapse └ Sub-archetype Difference-Split Collapse

telescoping-sumpartial-fractionsfraction-arithmetic identify-subproblems ↑ Prerequisites: fraction-arithmetic

📏 Long solution 💡 3 insights

Problem

Add up the fraction

\frac{1}{k^3 + 6k^2 + 8k}

, taking one term for every whole number k starting at 1 and going on forever. That infinite total equals a single fraction

\frac{a}{b}

, where a and b are positive integers with no common factor. Find a + b.

Pick an answer.

(A)

(B)

(C)

102

(D)

107

(E)

129

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Look for a Pattern

Factoring the denominator shows the three factors k, k+2, k+4 sit exactly 2 apart. Evenly spaced factors are the classic signal of a telescoping sum: rewrite each term as a difference of two block-fractions so that when you add many terms the middle cancels. Then only a few leftover pieces survive, and adding them gives the total.

1STEP 1

Factor the denominator

Pull the common factor k out of the cubic, then factor the leftover quadratic: $k(k+2)(k+4)$ .

k³ + 6k² + 8k = k(k² + 6k + 8) = k(k+2)(k+4)

2STEP 2

Spot the even spacing

The factors sit 2 apart, so subtracting neighboring block-fractions leaves the constant 4 on top of $k(k+2)(k+4)$ .

1/(k(k+2)) - 1/(k+2)(k+4) = ((k+4)-k)/(k(k+2)(k+4)) = 4/(k(k+2)(k+4))

3STEP 3

Name the block f(k)

Divide by 4 and name the block $f(k)=\frac{1}{k(k+2)}$ , so each term is $\frac{1}{4}(f(k)-f(k+2))$ .

1/(k(k+2)(k+4)) = 1/4(1/(k(k+2)) - 1/(k+2)(k+4)) = 1/4(f(k) - f(k+2)), f(k)=1/(k(k+2))

4STEP 4

Telescope the partial sum

Summing the first N terms cancels the middle, and the tail shrinks to 0, leaving $\frac{1}{4}(f(1)+f(2))$ .

Σ_k=1^N1/4(f(k)-f(k+2)) = 1/4(f(1)+f(2)-f(N+1)-f(N+2)) ⟶ 1/4(f(1)+f(2))

5STEP 5

Add the survivors

With $f(1)=\frac{1}{3}$ and $f(2)=\frac{1}{8}$ , one quarter of their sum is $\frac{11}{96}$ .

1/4(1/(1 · 3)+1/(2 · 4))=1/4(1/3+1/8)=1/4·11/24=11/96

6STEP 6

Reduce and add a + b

11 is prime and does not divide 96, so $\frac{11}{96}$ is already lowest terms and a + b = 107.

gcd(11,96)=1, a+b = 11+96 = 107

Answer

107

Estimate the sum with a few terms: 1/15 + 1/48 + 1/105 + ... is about 0.0667 + 0.0208 + 0.0095 + ... which piles up slowly toward roughly 0.1146, and 11/96 = 0.11458..., so the closed form is the right size. The surviving pieces 1/3 and 1/8 are exactly the two smallest-k blocks, which is what telescoping predicts.

💡Key takeaway

When factors are evenly spaced, rewrite each term as a difference so the middle cancels; only the first blocks survive, here giving (1/4)(1/3 + 1/8) = 11/96.

Factor the denominator
Spot the even spacing
Name the block f(k)
Telescope the partial sum
Add the survivors
Reduce and add a + b