1951 AMC 12 第 11 题

Question

一个无穷等比数列的和为

\dfrac{a}{1-r}

，其中

a

表示首项，

-1\lt r\lt1

表示公比。该数列各项平方之和为：

The limit of the sum of an infinite number of terms in a geometric progression is

\dfrac{a}{1-r},

where

a

denotes the first term and

-1\lt r\lt1

denotes the common ratio. The limit of the sum of their squares is:

\dfrac{a^2}{(1-r)^2}

\dfrac{a^2}{1+r^2}

\dfrac{a^2}{1-r^2}

\dfrac{4a^2}{1+r^2}

以上答案均不正确

None of these

将每一项平方后，公比也随之平方

Squaring every term also squares the common ratio

平方后的数列之和以

a^2+a^2r^2+a^2r^4+\cdots

开始

The squared series begins

a^2+a^2r^2+a^2r^4+\cdots

各项平方后仍组成等比数列，首项为

a^2

，公比为

r^2

。因为

|r|\lt1

，所以其和为

\frac{a^2}{1-r^2}\text{。}

因此，正确答案是 C。

The squared terms form a geometric series with first term

a^2

and ratio

r^2.

Since

|r|\lt1,

its sum is

\frac{a^2}{1-r^2}.

Thus, the correct answer is C.

其他年份的第 11 题