AMC10 2025 B

AMC10 2025 B · Q15

AMC10 2025 B · Q15. It mainly tests Rational expressions, Algebra misc.

The sum \[\sum_{k=1}^{\infty} \frac{1}{k^3 + 6k^2 + 8k}\] can be expressed as $\frac{a}{b}$, where a and b are relatively prime positive integers. What is a+b?

和 \[\sum_{k=1}^{\infty} \frac{1}{k^3 + 6k^2 + 8k}\] 可表示为$\frac{a}{b}$，其中$a$和$b$为互质正整数。求$a+b$？

(A) 89 89

(B) 97 97

(D) 107 107

(E) 129 129

Answer

Correct choice: (D)

正确答案：（D）

Solution

Through partial sum decomposition we obtain \[\frac{1}{k^3 + 6k^2 + 8k} = \frac{1}{8} \left( \frac{1}{k} - \frac{2}{k+2} + \frac{1}{k+4} \right),\] so one can write the expression as 18∑k=1∞(1k−2k+2+1k+4)=18∑k=1∞(1k−1k+2−1k+2+1k+4)=18∑k=1∞(1k−1k+2)+18∑k=1∞(1k+4−1k+2)=18∑k=1∞(1k−1k+2)−18∑k=1∞(1k+2−1k+4)=18∑k=1∞(1k−1k+2)−18∑k=3∞(1k−1k+2)=18∑k=12(1k−1k+2) Simplifying the sum and multiplying results in $\frac{11}{96}$, in which we finally have $11 + 96 =$ $\boxed{\textbf{(D)}~107}$ Note: For more info on Partial Sum Decompositions; AoPS Intermediate Algebra Chapter 17.2 (Telescoping) gives a brief introduction to them, which is sufficient enough to solve this problem.

通过部分分式分解，我们得到 \[\frac{1}{k^3 + 6k^2 + 8k} = \frac{1}{8} \left( \frac{1}{k} - \frac{2}{k+2} + \frac{1}{k+4} \right),\] 因此可以将表达式写成 $\frac{1}{8}\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{2}{k+2}+\frac{1}{k+4}\right)=\frac{1}{8}\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{k+2}-\frac{1}{k+2}+\frac{1}{k+4}\right)=\frac{1}{8}\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{k+2}\right)+\frac{1}{8}\sum_{k=1}^\infty\left(\frac{1}{k+4}-\frac{1}{k+2}\right)=\frac{1}{8}\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{k+2}\right)-\frac{1}{8}\sum_{k=1}^\infty\left(\frac{1}{k+2}-\frac{1}{k+4}\right)=\frac{1}{8}\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{k+2}\right)-\frac{1}{8}\sum_{k=3}^\infty\left(\frac{1}{k}-\frac{1}{k+2}\right)=\frac{1}{8}\sum_{k=1}^{2}\left(\frac{1}{k}-\frac{1}{k+2}\right)$ 化简和后乘以系数得到$\frac{11}{96}$，从而$11 + 96 = \boxed{\textbf{(D)}~107}$ 备注：关于部分分式分解的更多信息；AoPS中级代数第17.2章（伸缩和）对此有简要介绍，足以解此题。

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