2019 AMC 8 第 17 题

Question

下列乘积的值是多少？

\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?

\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*}

\displaystyle \dfrac{1}{2}

\displaystyle \dfrac{50}{99}

\displaystyle \dfrac{9800}{9801}

\displaystyle \dfrac{100}{99}

50

改写每个因子，使可以相消的部分对齐

Rewrite each factor to line up cancellations

最后只剩开头的

\frac12

和结尾的

\frac{100}{99}

Only the first

\frac12

and the last

\frac{100}{99}

remain

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可以把所有因子重新组合为

\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*}\text{。}

由这个形式可见，中间各项全部相消，只剩

\dfrac{1}{2} \cdot \dfrac{100}{99} = \dfrac{50}{99}\text{。}

正确答案是 B。

We can regroup all the factors as follows.

\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

\dfrac{1}{2} \cdot \dfrac{100}{99} = \dfrac{50}{99}.

Thus, the correct answer is B.

其他年份的第 17 题