2011 AIME II Problem 5

Attempt Problem 5 of the 2011 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AIME II solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

5.

The sum of the first 20112011 terms of a geometric series is 200.200. The sum of the first 40224022 terms of the same series is 380.380. Find the sum of the first 60336033 terms of the series.

Answer: 542
Concepts:geometric sequencesummation
Difficulty rating: 1970
Solution:

Group the series into blocks of 20112011 consecutive terms. Each term of the second block is r2011r^{2011} times the corresponding term of the first block, so the block sums form a geometric sequence with ratio r2011.r^{2011}. The first block sums to 200200 and the second block sums to 380200=180,380 - 200 = 180, so r2011=180200=910.r^{2011} = \frac{180}{200} = \frac{9}{10}.

The third block then sums to 180910=162,180 \cdot \frac{9}{10} = 162, so the sum of the first 60336033 terms is 380+162=542.380 + 162 = 542.

← Problem 4#4
Full Exam

Problem 5 in Other Years