2004 AIME I Problems/Problem 7

Problem

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

Solution

Let our polynomial be $P(x)$ .

It is clear that the coefficient of $x$ in $P(x)$ is $-1 + 2 - 3 + \ldots + 14 - 15 = -8$ , so $P(x) = 1 -8x + Cx^2 + Q(x)$ , where $Q(x)$ is some polynomial divisible by $x^3$ .

Then $P(-x) = 1 + 8x + Cx^2 + Q(-x)$ and so $P(x)\cdot P(-x) = 1 + (2C - 64)x^2 + R(x)$ , where $R(x)$ is some polynomial divisible by $x^3$ .

However, we also know $P(x)\cdot P(-x) = (1 - x)(1 + x)(1 +2x)(1 - 2x) \cdots (1 - 15x)(1 + 15x)$ $= (1 - x^2)(1 - 4x^2)\cdots(1 - 225x^2)$ $= 1 - (1 + 4 + \ldots + 225)x^2 + R(x)$ .

Equating coefficients, we have $2C - 64 = -(1 + 4 + \ldots + 225) = -1240$ , so $-2C = 1176$ and $|C| = \boxed{588}$ .

2004 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

2004 AIME I Problems/Problem 7

Problem

Solution

See also