1975 AMC 12 第 16 题

先试着解答 1975 AMC 12 第 16 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1975 AMC 12 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

16.

一个无穷等比级数的首项为正整数,公比为正整数的倒数,且级数之和为 33,则其前两项之和为
If the first term of an infinite geometric series is a positive integer, the common ratio is the reciprocal of a positive integer, and the sum of the series is 3,3, then the sum of the first two terms of the series is
13\frac13
23\frac23
83\frac83
22
92\frac92
答案:C
知识点:等比数列整除性求和
小提示:
将首项记为 aa,公比记为 1n\frac{1}{n}
Write the first term as aa and the ratio as 1n\frac{1}{n}
大提示:
从 a1−1n=3\frac{a}{1-\frac{1}{n}}=3 出发,利用 a,na,n 为正整数且 n>1n\gt1
From a1−1n=3,\frac{a}{1-\frac{1}{n}}=3, use that a,na,n are positive integers and n>1n\gt1
解答:
设首项为 aa,公比为 1n\frac{1}{n}。由级数收敛知 n>1n\gt1,且 a1−1n=3,a=3−3n。 \frac{a}{1-\frac{1}{n}}=3,\qquad a=3-\frac3n\text{。} 因而 nn 整除 33,所以 n=3n=3、a=2a=2。前两项之和为 2+23=832+\frac{2}{3}=\frac{8}{3}。
因此,正确答案是 C。
Let the first term be aa and the ratio 1n.\frac{1}{n}. Convergence gives n>1,n\gt1, and a1−1n=3,a=3−3n. \frac{a}{1-\frac{1}{n}}=3,\qquad a=3-\frac3n. Thus nn divides 3,3, so n=3n=3 and a=2.a=2. The first two terms sum to 2+23=83.2+\frac{2}{3}=\frac{8}{3}.
Therefore, the correct answer is C.
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