2011 AIME II Problem 5

Below is the professionally curated solution for Problem 5 of the 2011 AIME II, 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.

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Concepts:geometric sequencesummation

Difficulty rating: 1970

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.

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.

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