AMC12 2016 B

AMC12 2016 B · Q14

AMC12 2016 B · Q14. It mainly tests Functions basics, Manipulating equations.

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 $S$?

一个无穷等比数列的和是一个正数 $S$，且该数列的第二项为 1。问 $S$ 的最小可能值是多少？

(A) \frac{1+\sqrt{5}}{2} \frac{1+\sqrt{5}}{2}

(B) 2 2

(D) 3 3

(E) 4 4

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Let $r$ be the common ratio of the geometric series; then $S=\dfrac{1}{r}+1+r+r^2+\cdots=\dfrac{\frac{1}{r}}{1-r}=\dfrac{1}{r-r^2}.$ Because $S>0$, the smallest value of $S$ occurs when the value of $r-r^2$ is maximized. The graph of $f(r)=r-r^2$ is a downward-opening parabola with vertex $\left(\dfrac{1}{2},\dfrac{1}{4}\right)$, so the smallest possible value of $S$ is $\dfrac{1}{\left(\frac{1}{4}\right)}=4$. The optimal series is $2,1,\dfrac{1}{2},\dfrac{1}{4},\cdots$.

答案（E）：设 $r$ 为该等比数列的公比，则 $S=\dfrac{1}{r}+1+r+r^2+\cdots=\dfrac{\frac{1}{r}}{1-r}=\dfrac{1}{r-r^2}.$ 因为 $S>0$，当 $r-r^2$ 取最大值时，$S$ 取得最小值。函数 $f(r)=r-r^2$ 的图像是一条开口向下的抛物线，顶点为 $\left(\dfrac{1}{2},\dfrac{1}{4}\right)$，因此 $S$ 的最小可能值为 $\dfrac{1}{\left(\frac{1}{4}\right)}=4$。最优的级数为 $2,1,\dfrac{1}{2},\dfrac{1}{4},\cdots$。

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