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2004 AIME I Problems/Problem 7

Revision as of 22:21, 5 October 2020 by Jackshi2006 (talk | contribs) (→‎Solution 5)

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

Solution 6

This is just another way of summing the subsets of 2 from $[-1, 2, -3, 4, -5, 6, -7, 8, -9, 10, -11, 12, -13, 14, -15]$. Start from the right and multiply -15 to everything on its left. Use the distributive property and add all the 14 integers together to get 7. This gives us $-15 * 7$. Doing this for 14 gives us $14 * -7$, and for -13 we get $-13 * 6$. This pattern repeats where every two integers will multiple 7, 6,... to 0. Combining and simplifying the pattern give us this: $-(29 * 7 + 25 * 6 + 21 * 5 + 17 * 4 + 13 * 3 + 9 * 2 + 5*1)$. The expression gives us -588, or $C = \boxed{588}$. This is a good solution because it guarantees we never add a product twice, and the pattern is simple to add by hand.

-jackshi2006

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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