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

Revision as of 15:38, 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 $C = \sum_{1 \le i \neq j}^{15} S_iS_j$ . 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, $|C| = \left|\frac{64-1240}{2}\right| = \boxed{588}$ .

@@ Line 16: / Line 16: @@
 === Solution 2 ===
-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
+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_{1 \le i \neq j}^{15} S_iS_j</math>. Also, we know that
 <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>
 where the left-hand sum can be computed from:
 <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>
-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>.
+and the right-hand sum comes from the formula for the sum of the first <math>n</math> perfect squares. Therefore, <math>|C| = \left|\frac{64-1240}{2}\right| = \boxed{588}</math>.
 == 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
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