2025 AMC 10B Problem 15

1. Factor the bottom.
$$k^3 + 6k^2 + 8k = k(k^2 + 6k + 8) = k(k+2)(k+4).$$
The first term is $\frac{1}{1 \cdot 3 \cdot 5}$, the next is $\frac{1}{2 \cdot 4 \cdot 6}$, and so on.

2. Split each term into a difference. Sums like this are usually solved by writing each term as a difference, so that neighbouring terms cancel. Try
$$\frac{1}{k(k+2)} - \frac{1}{(k+2)(k+4)}.$$
Over the common bottom $k(k+2)(k+4)$, the top is $(k+4) - k = 4$. So this difference is $\frac{4}{k(k+2)(k+4)}$, which is $4$ times our term. Therefore
$$\frac{1}{k(k+2)(k+4)} = \frac14\left(\frac{1}{k(k+2)} - \frac{1}{(k+2)(k+4)}\right).$$

3. Write out the first few terms and watch them cancel. Leaving out the $\frac14$ for now:
$$\begin{aligned} k=1:&\quad \tfrac{1}{1\cdot3} - \tfrac{1}{3\cdot5}\\ k=2:&\quad \tfrac{1}{2\cdot4} - \tfrac{1}{4\cdot6}\\ k=3:&\quad \tfrac{1}{3\cdot5} - \tfrac{1}{5\cdot7}\\ k=4:&\quad \tfrac{1}{4\cdot6} - \tfrac{1}{6\cdot8} \end{aligned}$$
The $-\frac{1}{3\cdot 5}$ from $k = 1$ cancels with the $+\frac{1}{3\cdot5}$ from $k = 3$. The $-\frac{1}{4\cdot6}$ from $k=2$ cancels with $k=4$. Every negative piece is cancelled two rows later. The only positive pieces never cancelled are the first two, $\frac{1}{1\cdot3}$ and $\frac{1}{2\cdot4}$.

1. What about the far end? The pieces still waiting to be cancelled get smaller and smaller, closer and closer to $0$, so in the infinite sum they disappear.

5. Add up.
$$\frac14\left(\frac13 + \frac18\right) = \frac14 \cdot \frac{8 + 3}{24} = \frac{11}{96}.$$
$11$ and $96$ share no factor, so $a + b = 11 + 96 = 107$.

The answer is $\boxed{\textbf{(D)}\ 107}$.

When the bottom is a product of evenly spaced numbers, split each term into "first pair minus last pair" so that most pieces cancel. This is called a telescoping sum. The number in front ($\frac14$) is the gap between the outside factors, $k+4$ and $k$. Because the pieces here cancel two rows apart rather than one, two pieces survive at the start, and writing out four rows is the safe way to see that.

Split into three single fractions: $\frac{1}{k(k+2)(k+4)} = \frac{1}{8k} - \frac{1}{4(k+2)} + \frac{1}{8(k+4)}$. (Check at $k = 1$: $\frac18 - \frac1{12} + \frac1{40} = \frac{15 - 10 + 3}{120} = \frac{1}{15}$.) Adding over all $k$, almost everything cancels, but you have to track by hand which of $1, \frac12, \frac13, \frac14$ are left over, and that is easy to get wrong. It leads to the same $\frac{11}{96}$.

Factor the denominator:
$$k^3 + 6k^2 + 8k = k(k^2 + 6k + 8) = k(k+2)(k+4).$$

The three factors are spaced two apart, so pair the outer two off against each other. Write $g(k) = \frac{1}{k(k+2)}$ and note
$$g(k) - g(k+2) = \frac{1}{k(k+2)} - \frac{1}{(k+2)(k+4)} = \frac{(k+4) - k}{k(k+2)(k+4)} = \frac{4}{k(k+2)(k+4)}.$$
Hence each term of the sum is $\tfrac14\bigl(g(k) - g(k+2)\bigr)$, and the partial sum through $k = N$ telescopes with a step of two:
$$\sum_{k=1}^{N} \frac{1}{k(k+2)(k+4)} = \frac14\Bigl(g(1) + g(2) - g(N+1) - g(N+2)\Bigr).$$
Two terms survive at the front because the step is $2$, not $1$: $g(1)$ is never cancelled by anything, and neither is $g(2)$.

As $N \to \infty$ both $g(N+1)$ and $g(N+2)$ tend to $0$, so
$$\sum_{k=1}^{\infty} \frac{1}{k(k+2)(k+4)} = \frac14\left(\frac{1}{1\cdot 3} + \frac{1}{2\cdot 4}\right) = \frac14\left(\frac13 + \frac18\right) = \frac14 \cdot \frac{11}{24} = \frac{11}{96}.$$

Since $\gcd(11, 96) = 1$, we have $a + b = 11 + 96 = 107$.

The answer is $\boxed{\textbf{(D)}\ 107}$.