AMC 10 · 2025 · #15
Grade 6 algebraThe sum
k=1∑∞k3+6k2+8k1
can be expressed as ba, where a and b are relatively prime positive integers. What is a+b?
Pick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Add up the fraction $\frac{1}{k^3 + 6k^2 + 8k}$ for every whole number k starting at 1 and going on forever. The total equals some fraction $\frac{a}{b}$ in lowest terms. Find a + b.
Givens: The sum runs over all integers k from 1 to infinity; Each term is $\frac{1}{k^3 + 6k^2 + 8k}$; The total can be written as $\frac{a}{b}$ with a and b relatively prime positive integers
Unknowns: The value of a + b, where $\frac{a}{b}$ is the reduced total of the infinite sum
Understand
Restated: Add up the fraction $\frac{1}{k^3 + 6k^2 + 8k}$ for every whole number k starting at 1 and going on forever. The total equals some fraction $\frac{a}{b}$ in lowest terms. Find a + b.
Givens: The sum runs over all integers k from 1 to infinity; Each term is $\frac{1}{k^3 + 6k^2 + 8k}$; The total can be written as $\frac{a}{b}$ with a and b relatively prime positive integers
Plan
Primary tool: #5 Look for a Pattern
Secondary: #7 Identify Subproblems, #4 Introduce a Variable, #9 Solve an Easier Related Problem, #3 Eliminate Possibilities
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.
Execute — Answer: D
6.EE.A.3 Step 1 Factor the denominator
- Pull the common factor k out of the cubic, then factor the leftover quadratic.
- This turns the messy cubic into a clean product of three linear factors.
💡 A product of simple factors is far easier to work with than a cubic, and it exposes the spacing between the factors.
5.NF.A.1 Step 2 Spot the even spacing
- The factors k, k+2, k+4 are each 2 apart.
- Subtracting two neighboring block-fractions gives a numerator of exactly (k+4) - k = 4 over the full triple product, so one term equals one quarter of that difference.
💡 When factors are evenly spaced, a difference of neighboring products collapses to a constant on top, which is exactly what telescoping needs.
6.EE.A.2 Step 3 Name the block f(k)
- Divide the identity by 4 to write each term of the sum as a scaled difference.
- Let f(k) = $\frac{1}{k(k+2)}$ so every term becomes one quarter of f(k) minus f(k+2).
💡 Giving the repeated block a single name makes the cancellation pattern easy to see.
4.OA.C.5 Step 4 Telescope the partial sum
- Add the first N terms.
- Each -f(k+2) is cancelled by a later +f(k), so everything in the middle vanishes and only the first two f-values and the last two survive.
- As N grows, the last two shrink to 0 because f(k) heads to 0.
💡 A telescoping sum is a chain where each link cancels the next, leaving only the two ends.
5.NF.B.4 Step 5 Add the survivors
- Plug in f(1) = $\frac{1}{1*3}$ = $\frac{1}{3}$ and f(2) = $\frac{1}{2*4}$ = $\frac{1}{8}$.
- Add these unlike fractions, then multiply by the $\frac{1}{4}$ out front.
💡 Only the two starting blocks matter, so the whole infinite sum reduces to one short fraction calculation.
6.NS.B.4 Step 6 Reduce and add a + b
- Check that $\frac{11}{96}$ is already in lowest terms: 11 is prime and does not divide 96, so gcd(11, 96) = 1.
- Then a = 11 and b = 96, giving a + b = 107, which matches choice (D).
💡 Confirming the fraction cannot reduce guarantees a and b are the exact numbers the problem asks to add.
6.EE.A.3 Pull the common factor k out of the cubic, then factor the leftover quadratic. T 5.NF.A.1 The factors k, k+2, k+4 are each 2 apart. Subtracting two neighboring block-frac 6.EE.A.2 Divide the identity by 4 to write each term of the sum as a scaled difference. L 4.OA.C.5 Add the first N terms. Each -f(k+2) is cancelled by a later +f(k), so everything 5.NF.B.4 Plug in f(1) = $\frac{1}{1*3}$ = $\frac{1}{3}$ and f(2) = $\frac{1}{2*4}$ = $\fr 6.NS.B.4 Check that $\frac{11}{96}$ is already in lowest terms: 11 is prime and does not Review
Reasonableness: Estimate the sum with a few terms: $\frac{1}{15}$ + $\frac{1}{48}$ + $\frac{1}{105}$ + ... is about 0.0667 + 0.0208 + 0.0095 + ... which piles up slowly toward roughly 0.1146, and $\frac{11}{96}$ = 0.11458..., so the closed form is the right size. The surviving pieces $\frac{1}{3}$ and $\frac{1}{8}$ are exactly the two smallest-k blocks, which is what telescoping predicts.
Alternative: Split the term with full partial fractions: $\frac{1}{k(k+2)(k+4)}$ = $\frac{\frac{1}{8}}{k}$ - $\frac{\frac{1}{4}}{k+2}$ + $\frac{\frac{1}{8}}{k+4}$. Grouping the surviving $\frac{1}{k}$ pieces after cancellation gives ($\frac{1}{8}$)(1 + $\frac{1}{2}$) - ($\frac{1}{8}$)(2)($\frac{1}{3}$ + $\frac{1}{4}$ combined)... which collapses to the same $\frac{11}{96}$, confirming the block method.
CCSS standards used (min grade 6)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Factoring the cubic denominator into k(k+2)(k+4))5.NF.A.1Add and subtract fractions with unlike denominators (Subtracting two block-fractions to find the constant numerator 4)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Naming the repeated block f(k) = $\frac{1}{k(k+2)}$ to expose the pattern)4.OA.C.5Generate a number or shape pattern that follows a given rule (Tracking the telescoping cancellation across the partial sum)5.NF.B.4Apply and extend previous understandings of multiplication to multiply fractions (Adding $\frac{1}{3}$ + $\frac{1}{8}$ and multiplying by $\frac{1}{4}$ to get $\frac{11}{96}$)6.NS.B.4Find the greatest common factor of two whole numbers (Confirming gcd(11, 96) = 1 so a = 11 and b = 96)
⭐ When factors are evenly spaced, rewrite each term as a difference so the middle cancels; only the first blocks survive, here giving ($\frac{1}{4}$)($\frac{1}{3}$ + $\frac{1}{8}$) = $\frac{11}{96}$.
⭐ When factors are evenly spaced, rewrite each term as a difference so the middle cancels; only the first blocks survive, here giving ($\frac{1}{4}$)($\frac{1}{3}$ + $\frac{1}{8}$) = $\frac{11}{96}$.
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