Question

一个等比数列前 $2011$ 项的和为 $200$ 。同一个数列前 $4022$ 项的和为 $380$ 。求该数列前 $6033$ 项的和。

The sum of the first $2011$ terms of a geometric series is $200.$ The sum of the first $4022$ terms of the same series is $380.$ Find the sum of the first $6033$ terms of the series.

答案

答案：542

知识点：等比数列求和

难度评级：1970

小提示：

第 $2012$ 项到第 $4022$ 项中的每一项，都是前 $2011$ 项中对应项的 $r^{2011}$ 倍。

Terms $2012$ through $4022$ are $r^{2011}$ times the first $2011$ terms

大提示：

每一块连续 $2011$ 项的和都是前一块的 $r^{2011}$ 倍；从两个给定和中求出这个比值。

Each successive block of $2011$ terms is the previous block times $r^{2011};$ find that ratio from the two given sums

解答：

将数列分成每块 $2011$ 个连续项。第二块的每一项都是第一块对应项的 $r^{2011}$ 倍，所以各块和构成公比为 $r^{2011}$ 的等比数列。第一块和为 $200$ ，第二块和为 $380 - 200 = 180$ ，所以 $r^{2011} = \frac{180}{200} = \frac{9}{10}$ 。

第三块的和为 $180 \cdot \frac{9}{10} = 162$ ，所以前 $6033$ 项的和为 $380 + 162 = 542$ 。

Group the series into blocks of $2011$ consecutive terms. Each term of the second block is $r^{2011}$ times the corresponding term of the first block, so the block sums form a geometric sequence with ratio $r^{2011}.$ The first block sums to $200$ and the second block sums to $380 - 200 = 180,$ so $r^{2011} = \frac{180}{200} = \frac{9}{10}.$

The third block then sums to $180 \cdot \frac{9}{10} = 162,$ so the sum of the first $6033$ terms is $380 + 162 = 542.$

2011 AIME II 第 5 题

答案：542

难度评级：1970

小提示：

大提示：

解答：

其他年份的第 5 题