2022 AMC 8 第 8 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。

8.

下列表达式的值是多少?

132435182019212022 \dfrac{1}{3} \cdot \dfrac{2}{4} \cdot \dfrac{3}{5} \cdots \dfrac{18}{20} \cdot \dfrac{19}{21} \cdot \dfrac{20}{22}

What is the value of:

132435182019212022 \dfrac{1}{3} \cdot \dfrac{2}{4} \cdot \dfrac{3}{5} \cdots \dfrac{18}{20} \cdot \dfrac{19}{21} \cdot \dfrac{20}{22}

1462\displaystyle \dfrac{1}{462}

1231\displaystyle \dfrac{1}{231}

1132\displaystyle \dfrac{1}{132}

2213\displaystyle \dfrac{2}{213}

122\displaystyle \dfrac{1}{22}

答案:B
知识点:裂项相消分数
难度评级:1020
解答:

332020 的每个整数都既作为分母出现一次,也作为分子出现一次,所以全部相消。

相消后,分子只剩 1122,分母只剩 21212222

剩下的分数是 122122\dfrac{1 \cdot 2}{21 \cdot 22} ,化简为 2462=1231\dfrac{2}{462} = \dfrac{1}{231}

正确答案是 B

Since every integer from 33 to 2020 occurs once as a denominator and once as a numerator, they cancel each other out.

After canceling every number out, we have only 11 and 22 left as numerators and 2121 and 2222 left as denominators.

The remaining fraction is 122122. \dfrac{1 \cdot 2}{21 \cdot 22} . This simplifies to 2462=1231 \dfrac{2}{462} = \dfrac{1}{231}

Thus, the correct answer is B.

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