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

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== Problem ==
 
== Problem ==
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>
  
== Solution ==
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__TOC__
Let our [[polynomial]] be <math>P(x)</math>.  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>.  However, we also know <math>P(x)\cdot P(-x) = (1 - x)(1 + x)(1 +2x)(1 - 2x) \cdots (1 - 15x)(1 + 15x)</math> <math>= (1 - x^2)(1 - 4x^2)\cdots(1 - 225x^2)</math> <math>= 1 - (1 + 4 + \ldots + 225)x^2 + R(x)</math>.
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== Solutions ==
Equating coefficients, we have <math>2C - 64 = -(1 + 4 + \ldots + 225) = -1240</math>, so <math>-2C = 1176</math> and <math>|C| = 588</math>.
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=== Solution 1 ===
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Let our [[polynomial]] be <math>P(x)</math>.
  
== See also ==
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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>.
* [[2004 AIME I Problems/Problem 6| Previous problem]]
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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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However, we also know <math>P(x)\cdot P(-x) = (1 - x)(1 + x)(1 +2x)(1 - 2x) \cdots (1 - 15x)(1 + 15x)</math> <math>= (1 - x^2)(1 - 4x^2)\cdots(1 - 225x^2)</math> <math>= 1 - (1 + 4 + \ldots + 225)x^2 + R(x)</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>.
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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_{1 \le i \neq j}^{15} S_iS_j</math>. Also, we know that
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<cmath>\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*}</cmath>
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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| = \left|\frac{64-1240}{2}\right| = \boxed{588}</math>.
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=== Solution 3 (Bash)===
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Consider the set <math>[-1, 2,-3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15]</math>. Denote by <math>S</math> all size 2 subsets of this set. Replace each element of <math>S</math> by the product of the elements. Now, the quantity we seek is the sum of each element. Since consecutive elements add to <math>1</math> or <math>-1</math>, we can simplify this to <math>588</math>
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==Solution 4==
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Let set <math>N</math> be <math>\{-1, -3, \ldots -15\}</math> and set <math>P</math> be <math>\{2, 4, \ldots 14\}</math>. The sum of the negative <math>x^2</math> coefficients is the sum of the products of the elements in all two element sets such that one element is from <math>N</math> and the other is from <math>P</math>. Each summand is a term in the expansion of
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<cmath>(-1 - 3 - \ldots - 15)(2 + 4 + \ldots + 14)</cmath>
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which equals <math>-56 * 64 = -(60^2 - 4^2) = -3584</math>. The sum of the positive <math>x^2</math> coefficients is the sum of the products of all two element sets such that the two elements are either both in <math>N</math> or both in <math>P</math>. By counting, the sum is <math>2992</math>, so the sum of all <math>x^2</math> coefficients is <math>-588</math>. Thus, the answer is <math>\boxed{588}</math>.
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== Solution 5==
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We can find out the coefficient of <math>x^2</math> by multiplying every pair of two coefficients for <math>x</math>. This means that we multiply <math>-1</math> by <math>2,-3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15</math> and <math>2</math> by <math>3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15</math>. and etc. This sum can be easily simplified and is equal to <math>(-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)</math> or <math>588</math>.
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-David Camacho
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==Solution 6==
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This is just another way of summing the subsets of 2 from <math>[-1, 2, -3, 4, -5, 6, -7, 8, -9, 10, -11, 12, -13, 14, -15]</math>. 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 <math>-15 * 7</math>. Doing this for 14 gives us <math>14 * -7</math>, and for -13 we get <math>-13 * 6</math>. This pattern repeats where every two integers will multiple 7, 6,... to 0. Combining and simplifying the pattern give us this: <math>-(29 * 7 + 25 * 6 + 21 * 5 + 17 * 4 + 13 * 3 + 9 * 2 + 5*1)</math>. The expression gives us -588, or <math>C = \boxed{588}</math>. This is a good solution because it guarantees we never add a product twice, and the pattern is simple to add by hand.
  
* [[2004 AIME I Problems/Problem 8| Next problem]]
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-jackshi2006
  
* [[2004 AIME I Problems]]
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== See also ==
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{{AIME box|year=2004|n=I|num-b=6|num-a=8}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Revision as of 19:27, 9 May 2021

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 $588$

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 (ProblemsAnswer KeyResources)
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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