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 $S?$

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

$\ 2$

$\ \sqrt{5}$

$\ 3$

$\ 4$

答案：E

知识点：等比数列配方法最优化

难度评级：1540

解答：

设首项为 $a$ 公比为 $r$ 由第二项为一可得因此要使级数和最小，就要找使下式最大的 $r$ ：这在 $r=0.5$ 时成立。 $\begin{aligned}S&=\dfrac{a}{1-r}\\ &= \dfrac{ar}{r(1-r)} \\&= \dfrac{1}{r(1-r)}.\end{aligned}$ $r(1-r)= 0.25-(r-0.5)^2.$

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 .

2016 AMC 10B 第 16 题

答案：E

难度评级：1540

解答：

其他年份的第 16 题