Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2004 AIME I Problems/Problem 7"
m (→‎See also)
Line 7: Line 7:
  
 
== See also ==
 
== See also ==
 +
* [[2004 AIME I Problems/Problem 6| Previous problem]]
 +
 +
* [[2004 AIME I Problems/Problem 8| Next problem]]
 +
 
* [[2004 AIME I Problems]]
 
* [[2004 AIME I Problems]]
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 14:59, 18 August 2006

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 $|C| = 588$.

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_7&oldid=9690"