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Difference between revisions of "2004 AIME I Problems/Problem 7"

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(slightly different solution)
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== Problem ==
 
== Problem ==
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Let <math> C </math> be the coefficient of <math> x^2 </math> in the expansion of the product <math> (1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x). </math> Find <math> |C|. </math>
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Let <math> C </math> be the [[coefficient]] of <math> x^2 </math> in the expansion of the product <math> (1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x). </math> Find <math> |C|. </math>
  
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__TOC__
 
== Solution ==
 
== Solution ==
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=== Solution 1 ===
 
Let our [[polynomial]] be <math>P(x)</math>.
 
Let our [[polynomial]] be <math>P(x)</math>.
  
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It is clear that the [[coefficient]] of <math>x</math> in <math>P(x)</math> is <math>-1 + 2 - 3 + \ldots + 14 - 15 = -8</math>, so <math>P(x) = 1 -8x + Cx^2 + Q(x)</math>, where <math>Q(x)</math> is some polynomial [[divisibility | divisible]] by <math>x^3</math>.
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It is clear that the coefficient of <math>x</math> in <math>P(x)</math> is <math>-1 + 2 - 3 + \ldots + 14 - 15 = -8</math>, so <math>P(x) = 1 -8x + Cx^2 + Q(x)</math>, where <math>Q(x)</math> is some polynomial [[divisibility | divisible]] by <math>x^3</math>.
  
 
Then <math>P(-x) = 1 + 8x + Cx^2 + Q(-x)</math> and so <math>P(x)\cdot P(-x) = 1 + (2C - 64)x^2 + R(x)</math>, where <math>R(x)</math> is some polynomial divisible by <math>x^3</math>.
 
Then <math>P(-x) = 1 + 8x + Cx^2 + Q(-x)</math> and so <math>P(x)\cdot P(-x) = 1 + (2C - 64)x^2 + R(x)</math>, where <math>R(x)</math> is some polynomial divisible by <math>x^3</math>.
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Equating coefficients, we have <math>2C - 64 = -(1 + 4 + \ldots + 225) = -1240</math>, so <math>-2C = 1176</math> and <math>|C| = \boxed{588}</math>.
 
Equating coefficients, we have <math>2C - 64 = -(1 + 4 + \ldots + 225) = -1240</math>, so <math>-2C = 1176</math> and <math>|C| = \boxed{588}</math>.
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=== Solution 2 ===
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Let <math>S</math> be the [[set]] of integers <math>\{-1,2,-3,\ldots,14,-15\}</math>. The coefficient of <math>x^2</math> in the expansion is equal to the sum of the product of each pair of distinct terms, or <math>C = \sum_{i \neq j \ge 1}^{15} S_iS_j</math>. Also, we know that
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<center><math>\begin{align*}\left(\sum_{i=1}^{n} S_i\right)^2 &= \left(\sum_{i=1}^{n} S_i^2\right) + 2\left(\sum_{1 \le i \neq j}^{15} S_iS_j\right)\\ (-8)^2 &= \frac{15(15+1)(2\cdot 15+1)}{6} + 2C\end{align*}</math></center>
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where the left-hand sum can be computed from:
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<center><math>\sum_{i=1}^{15} S_i = S_{15} + \left(\sum_{i=1}^{7} S_{2i-1} + S_{2i}\right) = -15 + 7 = -8</math></center>
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and the right-hand sum comes from the formula for the sum of the first <math>n</math> perfect squares. Therefore, <math>C = \frac{64-1240}{2} = -588</math>.
  
 
== See also ==
 
== See also ==

Revision as of 15:37, 27 April 2008

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

Solution 1

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 $|C| = \boxed{588}$.

Solution 2

Let $S$ be the set of integers $\{-1,2,-3,\ldots,14,-15\}$. The coefficient of $x^2$ in the expansion is equal to the sum of the product of each pair of distinct terms, or $C = \sum_{i \neq j \ge 1}^{15} S_iS_j$. Also, we know that

$\begin{align*}\left(\sum_{i=1}^{n} S_i\right)^2 &= \left(\sum_{i=1}^{n} S_i^2\right) + 2\left(\sum_{1 \le i \neq j}^{15} S_iS_j\right)\\ (-8)^2 &= \frac{15(15+1)(2\cdot 15+1)}{6} + 2C\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

where the left-hand sum can be computed from:

$\sum_{i=1}^{15} S_i = S_{15} + \left(\sum_{i=1}^{7} S_{2i-1} + S_{2i}\right) = -15 + 7 = -8$

and the right-hand sum comes from the formula for the sum of the first $n$ perfect squares. Therefore, $C = \frac{64-1240}{2} = -588$.

See also

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
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