2025 AMC 10B Problem 15

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 $\boxed{\textbf{(D)}\ 107}$.

A sum of reciprocals of a product of linear factors in arithmetic progression always telescopes: group the factors into a block and shift it. Here the common difference is $2$ and there are three factors, so the natural block is the product of two consecutive factors and the shift is by one place, i.e. by $2$ in $k$. The constant $\frac14$ is exactly the gap between the outermost factors, which is why the identity is worth deriving rather than memorising. The general shape, $\frac{1}{k(k+d)(k+2d)} = \frac{1}{2d}\left[\frac{1}{k(k+d)} - \frac{1}{(k+d)(k+2d)}\right]$, covers every problem of this type.

Full partial fractions: $\frac{1}{k(k+2)(k+4)} = \frac{1/8}{k} - \frac{1/4}{k+2} + \frac{1/8}{k+4}$. Summing to $N$ and cancelling the common harmonic tail leaves
$$\frac18\Bigl[\bigl(1 + \tfrac12\bigr)\cdot 2 - \bigl(1 + \tfrac12 + \tfrac13 + \tfrac14\bigr)\Bigr] + O\!\left(\tfrac1N\right) = \frac18\left(3 - \frac{25}{12}\right) = \frac18 \cdot \frac{11}{12} = \frac{11}{96}.$$
This route is more error-prone precisely because the surviving boundary terms must be tracked by hand.

• Telescoping with a step of 2 but letting only one term survive at the front; both 1/(13) and 1/(24) remain, and dropping the second gives 1/12 instead of 11/96.
• Factoring the cubic as $k(k+2)(k+4)$ correctly but forgetting the factor $\frac14$ in the telescoping identity, which multiplies the answer by four.
• Answering $\frac{11}{96}$ itself rather than $a + b$.