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

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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 $|C|.$

Solutions

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 $\begin{align*}\left(\sum_{i=1}^{n} S_i\right)^2 &= \left(\sum_{i=1}^{n} S_i^2\right) + 2\left(\sum_{1 \le i \neq j}^{15} S_iS_j\right)\\ (-8)^2 &= \frac{15(15+1)(2\cdot 15+1)}{6} + 2C\end{align*}$ 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, $|C| = \left|\frac{64-1240}{2}\right| = \boxed{588}$ .

Solution 3 (Bash)

Consider the set $[-1, 2,-3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15]$ . Denote by $S$ all size 2 subsets of this set. Replace each element of $S$ by the product of the elements. Now, the quantity we seek is the sum of each element. Since consecutive elements add to $1$ or $-1$ , we can simplify this to $|-1\cdot(-7)+2\cdot(-9)-3\cdot(-6)+4\cdot(-10)-5\cdot(-5)+\ldots+12\cdot(-14)-13\cdot(-1)+14\cdot(-15)|=|-588|=\boxed{588}$ .

Solution 4

Let set $N$ be $\{-1, -3, \ldots -15\}$ and set $P$ be $\{2, 4, \ldots 14\}$ . The sum of the negative $x^2$ coefficients is the sum of the products of the elements in all two element sets such that one element is from $N$ and the other is from $P$ . Each summand is a term in the expansion of $(-1 - 3 - \ldots - 15)(2 + 4 + \ldots + 14)$ which equals $-56 * 64 = -(60^2 - 4^2) = -3584$ . The sum of the positive $x^2$ coefficients is the sum of the products of all two element sets such that the two elements are either both in $N$ or both in $P$ . By counting, the sum is $2992$ , so the sum of all $x^2$ coefficients is $-588$ . Thus, the answer is $\boxed{588}$ .

Solution 5

We can find out the coefficient of $x^2$ by multiplying every pair of two coefficients for $x$ . This means that we multiply $-1$ by $2,-3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15$ and $2$ by $3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15$ . and etc. This sum can be easily simplified and is equal to $(-1)(-7)+(-3)(-6)+(-5)(-5)+(-7)(-4)+(-9)(-3)+(-11)(-2)+(-13)(-1)+2(-9)+4(-10)+6(-11)+8(-12)+10(-13)+12(-14)+14(-15)$ or $588$ .

-David Camacho

@@ Line 3: / Line 3: @@
 __TOC__
-== Solution ==
+== Solutions ==
 === Solution 1 ===
 Let our [[polynomial]] be <math>P(x)</math>.

2004 AIME I (Problems • Answer Key • Resources)
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