Question

对无穷级数 $\begin{aligned} &1-\dfrac12-\dfrac14+\dfrac18\\ &{}-\dfrac1{16}-\dfrac1{32}+\dfrac1{64}\\ &{}-\dfrac1{128}-\cdots \end{aligned}\text{，}$ 设其极限和为 $S$ 。则 $S$ 等于：

For the infinite series $\begin{aligned} &1-\dfrac12-\dfrac14+\dfrac18\\ &{}-\dfrac1{16}-\dfrac1{32}+\dfrac1{64}\\ &{}-\dfrac1{128}-\cdots, \end{aligned}$ let $S$ be the (limiting) sum. Then $S$ equals:

$0$

$\dfrac27$

$\dfrac67$

$\dfrac9{32}$

$\dfrac{27}{32}$

答案：B

知识点：等比数列配对与分组求和

难度评级：1590

小提示：

将各项依次每三项分成一组

Group the terms in consecutive blocks of three

大提示：

每一组都是前一组的 $\frac{1}{8}$

Each block is $\frac{1}{8}$ times the preceding block

解答：

将级数分组为 $\begin{aligned} S={}&\left(1-\frac12-\frac14\right)\\ &+\left(\frac18-\frac1{16}-\frac1{32}\right) +\cdots \end{aligned}\text{。}$ 第一组为 $\frac{1}{4}$ ，各组构成公比为 $\frac{1}{8}$ 的等比级数。因此 $S=\frac{\frac{1}{4}}{1-\frac{1}{8}}=\frac27\text{。}$

因此，正确答案是 B。

Group the series as $\begin{aligned} S={}&\left(1-\frac12-\frac14\right)\\ &+\left(\frac18-\frac1{16}-\frac1{32}\right) +\cdots. \end{aligned}$ The first block is $\frac{1}{4},$ and successive blocks form a geometric series with ratio $\frac{1}{8}.$ Hence $S=\frac{\frac{1}{4}}{1-\frac{1}{8}}=\frac27.$

Therefore, the correct answer is B.

1959 AMC 12 第 49 题

答案：B

难度评级：1590

小提示：

大提示：

解答：

其他年份的第 49 题