1979 AMC 12 第 6 题

Question

\frac32+\frac54+\frac98+\frac{17}{16}

+\frac{33}{32}+\frac{65}{64}-7=

-\frac1{64}

-\frac1{16}

0

\frac1{16}

\frac1{64}

把每个分数写成

1

加上

2

的某个负整数次幂

Write each fraction as

1

plus a reciprocal power of

2

把

\frac12+\frac14+\cdots+\frac1{64}

看作有限等比数列求和

Sum

\frac12+\frac14+\cdots+\frac1{64}

as a finite geometric series

每个分子都比分母大一，因此原式为

\begin{aligned} 6+\biggl(&\frac12+\frac14+\frac18\\ &+\frac1{16}+\frac1{32}+\frac1{64}\biggr)-7 \end{aligned}\text{。}

括号内的等比数列之和为

1-\frac{1}{64}=\frac{63}{64}

。所以原式的值为

6+\frac{63}{64}-7=-\frac{1}{64}

。

因此，正确答案是 A。

Each numerator is one more than its denominator, so the expression is

\begin{aligned} 6+\biggl(&\frac12+\frac14+\frac18\\ &+\frac1{16}+\frac1{32}+\frac1{64}\biggr)-7. \end{aligned}

The parenthesized geometric sum is

1-\frac{1}{64}=\frac{63}{64}.

Therefore the value is

6+\frac{63}{64}-7=-\frac{1}{64}.

Therefore, the correct answer is A.

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