2019 AMC 8 第 17 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

17.

下列乘积的值是多少?

(1322)(2433)(3544)(97999898)(981009999)\begin{align*}\displaystyle & \left(\dfrac{1\cdot 3}{2\cdot 2}\right)\left(\dfrac{2\cdot 4}{3\cdot 3}\right)\left(\dfrac{3\cdot 5}{4\cdot 4}\right) \\ & \dots\left(\dfrac{97\cdot 99}{98\cdot 98}\right)\left(\dfrac{98\cdot 100}{99\cdot 99}\right) \end{align*}

What is the value of the product below?

(1322)(2433)(3544)(97999898)(981009999)\begin{align*}\displaystyle & \left(\dfrac{1\cdot 3}{2\cdot 2}\right)\left(\dfrac{2\cdot 4}{3\cdot 3}\right)\left(\dfrac{3\cdot 5}{4\cdot 4}\right) \\ & \dots\left(\dfrac{97\cdot 99}{98\cdot 98}\right)\left(\dfrac{98\cdot 100}{99\cdot 99}\right) \end{align*}

12\displaystyle \dfrac{1}{2}

5099\displaystyle \dfrac{50}{99}

98009801\displaystyle \dfrac{9800}{9801}

10099\displaystyle \dfrac{100}{99}

5050

答案:B
知识点:裂项相消分数
难度评级:1310
视频讲解:
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文字解答:

可以把所有因子重新组合为 12(3223)(4334)(99989899)10099\begin{align*} & \dfrac{1}{2} \cdot \left(\dfrac{3 \cdot 2}{2\cdot 3}\right)\left(\dfrac{4\cdot 3}{3\cdot 4}\right) \\ & \dots\left(\dfrac{99\cdot 98}{98\cdot 99}\right) \cdot \dfrac{100}{99} \end{align*}

由这个形式可见,中间各项全部相消,只剩 1210099=5099. \dfrac{1}{2} \cdot \dfrac{100}{99} = \dfrac{50}{99}.

正确答案是 B

We can regroup all the factors as follows. 12(3223)(4334)(99989899)10099\begin{align*} & \dfrac{1}{2} \cdot \left(\dfrac{3 \cdot 2}{2\cdot 3}\right)\left(\dfrac{4\cdot 3}{3\cdot 4}\right) \\ & \dots\left(\dfrac{99\cdot 98}{98\cdot 99}\right) \cdot \dfrac{100}{99} \end{align*}

From this representation, we can see that all the middle terms cancel leaving only 1210099=5099. \dfrac{1}{2} \cdot \dfrac{100}{99} = \dfrac{50}{99}.

Thus, the correct answer is B.

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