2004 AIME I Problem 7

Below is the professionally curated solution for Problem 7 of the 2004 AIME I, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AIME I solutions, or check the answer key.

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Concepts:polynomialalgebraic manipulationsum of first n squares

Difficulty rating: 2390

7.

Let CC be the coefficient of x2x^2 in the expansion of the product (1x)(1+2x)(13x)(1+14x)(115x).(1 - x)(1 + 2x)(1 - 3x) \cdots (1 + 14x)(1 - 15x). Find C.|C|.

Solution:

Write the product as k=115(1+akx)\prod_{k=1}^{15} (1 + a_k x) with ak=(1)kk.a_k = (-1)^k k. An x2x^2 term arises by choosing the xx-term from two factors, so C=i<jaiaj,C = \sum_{i \lt j} a_i a_j, and C=(ak)2ak22.C = \frac{\left(\sum a_k\right)^2 - \sum a_k^2}{2}.

The alternating sum is (1+2)+(3+4)++(13+14)15=715=8,(-1 + 2) + (-3 + 4) + \cdots + (-13 + 14) - 15 = 7 - 15 = -8, and ak2=12+22++152=1516316=1240.\sum a_k^2 = 1^2 + 2^2 + \cdots + 15^2 = \frac{15 \cdot 16 \cdot 31}{6} = 1240.

Thus C=6412402=588,C = \frac{64 - 1240}{2} = -588, so C=588.|C| = 588.

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