1952 AMC 12 第 50 题

Question

一条初始长度为

1

英寸的线段按下列规律增长，其中第一项为初始长度。

\begin{gathered} 1+\frac14\sqrt2+\frac14+\frac1{16}\sqrt2\\ {}+\frac1{16}+\frac1{64}\sqrt2+\frac1{64}+\cdots \end{gathered}\text{。}

若增长过程无限继续，线段长度的极限为：

A line initially

1

inch long grows according to the following law, where the first term is the initial length.

\begin{gathered} 1+\frac14\sqrt2+\frac14+\frac1{16}\sqrt2\\ {}+\frac1{16}+\frac1{64}\sqrt2+\frac1{64}+\cdots. \end{gathered}

If the growth process continues forever, the limit of the length of the line is:

\infty

\dfrac43

\dfrac38

\dfrac13(4+\sqrt2)

\dfrac23(4+\sqrt2)

将含有相同

\frac{1}{4}

的幂的每一对项分为一组

Group each pair having the same power of

\frac{1}{4}

初始项

1

之后的各项之和等于

(1+\sqrt2)\sum_{k=1}^{\infty}4^{-k}

The terms after the initial

1

equal

(1+\sqrt2)\sum_{k=1}^{\infty}4^{-k}

在初始项之后，每个幂

4^{-k}

分别单独出现一次，并乘以

\sqrt2

出现一次。因为

\sum_{k=1}^{\infty}4^{-k}=\frac{1}{3}

，所以极限为

1+\frac{1+\sqrt2}{3}=\frac{4+\sqrt2}{3}\text{。}

因此，正确答案是 D。

After the initial term, each power

4^{-k}

appears once by itself and once multiplied by

\sqrt2.

Since

\sum_{k=1}^{\infty}4^{-k}=\frac{1}{3},

the limit is

1+\frac{1+\sqrt2}{3}=\frac{4+\sqrt2}{3}.

Thus, the correct answer is D.

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