2016 AMC 10B 第 16 题

Question

一个无穷等比级数的和是正数

S

，且该级数的第二项是

1

。

S

的最小可能值是多少？

The sum of an infinite geometric series is a positive number

S,

and the second term in the series is

1.

What is the smallest possible value of

S?

\ \dfrac{1+\sqrt{5}}{2}

\ 2

\ \sqrt{5}

\ 3

\ 4

设首项为

a

，公比为

r

。

Write the first term as

a

and common ratio as

r

条件

ar=1

使得

S=\frac{1}{r(1-r)}

。

The condition

ar=1

makes

S=\frac{1}{r(1-r)}

设首项为

a

，公比为

r

。由第二项为一可得

\begin{aligned}S&=\dfrac{a}{1-r}\\ &= \dfrac{ar}{r(1-r)} \\&= \dfrac{1}{r(1-r)}\end{aligned}\text{。}

因此要使级数和最小，就要找使下式最大的

r

：

r(1-r)= 0.25-(r-0.5)^2\text{。}

这个最大值在

r=0.5

时取到。

所以

S = \dfrac{1}{0.5(0.5)}= 4

。

所以正确答案是 E。

Let the first value of the series be

a,

and let the ratio be

r.

Thus,

\begin{aligned}S&=\dfrac{a}{1-r}\\ &= \dfrac{ar}{r(1-r)} \\&= \dfrac{1}{r(1-r)}.\end{aligned}

This means we have to find

r

that maximizes

r(1-r)= 0.25-(r-0.5)^2.

This maximization will happen with

r=0.5.

Therefore,

S = \dfrac{1}{0.5(0.5)}= 4.

Thus, the correct answer is E .

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