AIME 2004 I · 第 7 题

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

$$\begin{aligned}\left(\sum_{i=1}^{n} S_i\right)^2 &= \left(\sum_{i=1}^{n} S_i^2\right) + 2\left(\sum_{1 \le i \neq j}^{15} S_iS_j\right)\\ (-8)^2 &= \frac{15(15+1)(2\cdot 15+1)}{6} + 2C\end{aligned}$$ 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 $588$

Solution 4 (Easy)

We know that this polynomial has roots $1, -\frac{1}{2}, \frac{1}{3},\ldots$ and the coefficient of $x^2$ will be the sum of the product taken by $13$. However, since this is closer to the constant side, we can create a new polynomial with the reciprocal roots which will make $C$ the coefficient of $x^{13}$ and thus the sum of the reciprocal roots taken by 2. We can calculate this with $1\cdot(8-1)-2\cdot(8-(-2))+\ldots+15(8-15)$. This will give us $2C = -1176$ giving us a final answer of $|C| = \fbox{588}$.

~ Vedoral

Solution 4.5 (Easier Explanation)

In this polynomial, we can see that by Vieta’s Formulae, the coefficient of $x^2$ is the sum of the products of every 13 of the roots. Let us denote the roots as $r_1, r_2, \dots, r_{15}$.

Then, by Vieta’s, we have the sum:

$$r_1 r_2 \dots r_{13} + \dots + r_3 r_4 \dots r_{15}$$ Also, since we know that the product of the roots is $1$ (again by Vieta’s Formulae), we can rewrite an arbitrary term of this expression. For example, the term $r_1 r_2 \dots r_{13}$ is equal to:

$$\frac{r_1 r_2 \dots r_{15}}{r_{14} r_{15}} = \frac{1}{r_{14} r_{15}}$$ Now we are just finding the value of the expression, because this is just an expansion of:

$$\frac{\left(\frac{1}{r_1} + \dots + \frac{1}{r_{15}}\right)^2 - \left(\frac{1}{r_1^2} + \dots + \frac{1}{r_{15}^2}\right)}{2}$$ Which evaluates to:

$$8 \cdot \frac{147}{2} = \boxed{588}$$ ~eqb5000, minor latex fixes and spelling

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

Solution 7

We expand and obtain $\left(x-1\right)\left(2x-1\right)\left(3x-1\right)\cdots\left(15x-1\right) = 1307674368000x^{15} - 948550176000x^{14} - 689324826240x^{13} + 2733483288464x^{12} + 82808260416x^{11} - 23038684088x^{10} - 3811851848x^9 + 828730833x^8 + 81228128x^7 - 14661124x^6 - 853104x^5 + 132902x^4 + 4256x^3 - \boxed{588}x^2 - 8x + 1.$

Do not do this in an actual competition.

~Sliced_Bread

Solution 8 (Bash Yay!)

Notice a pattern betweenthe x cooeficient and the x^2

x {-1,1, -2, 2, -3, 3..}

And then we can bash out the x^2 terms absolute value:

x^2={0, -2, -5, -13,....} you get $\boxed{588}$

~Aarav22