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Difference between revisions of "1986 AIME Problems/Problem 11"

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== Problem ==
 
== Problem ==
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The polynomial <math>1-x+x^2-x^3+\cdots+x^{16}-x^{17}</math> may be written in the form <math>a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}</math>, where <math>\displaystyle y=x+1</math> and thet <math>\displaystyle a_i</math>'s are constants. Find the value of <math>\displaystyle a_2</math>.
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The [[polynomial]] <math>1-x+x^2-x^3+\cdots+x^{16}-x^{17}</math> may be written in the form <math>a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}</math>, where <math>\displaystyle y=x+1</math> and the <math>\displaystyle a_i</math>'s are [[constant]]s. Find the value of <math>\displaystyle a_2</math>.
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== Solution ==
 
== Solution ==
 
Since <math>(1+x)(1-x+x^2-x^3+\cdots +x^{16}-x^{17})=1-x^{18}</math>, we have
 
Since <math>(1+x)(1-x+x^2-x^3+\cdots +x^{16}-x^{17})=1-x^{18}</math>, we have
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<math>y(a_0+a_1y+a_2y^2+\cdots +a_{17}y^{17})=1-(y-1)^{18}</math>
 
<math>y(a_0+a_1y+a_2y^2+\cdots +a_{17}y^{17})=1-(y-1)^{18}</math>
  
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So <math>a_2</math> is the <math>y^3</math> [[coefficient]], which, by the [[Binomial Theorem]], is <math>\frac{18\cdot 17\cdot 16}{3\cdot 2\cdot 1}=3\cdot 17\cdot 16=816</math>
  
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So <math>a_2</math> is the <math>y^3</math> coefficient, which, by the Binomial Theorem, is <math>\frac{18\cdot 17\cdot 16}{3\cdot 2\cdot 1}=3\cdot 17\cdot 16=816</math>
 
 
== See also ==
 
== See also ==
 
{{AIME box|year=1986|num-b=10|num-a=12}}
 
{{AIME box|year=1986|num-b=10|num-a=12}}
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* [[American Invitational Mathematics Examination]]
 
* [[American Invitational Mathematics Examination]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
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[[Category:Intermediate Algebra Problems]]

Revision as of 19:21, 12 September 2007

Problem

The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$, where $\displaystyle y=x+1$ and the $\displaystyle a_i$'s are constants. Find the value of $\displaystyle a_2$.

Solution

Since $(1+x)(1-x+x^2-x^3+\cdots +x^{16}-x^{17})=1-x^{18}$, we have

$y(a_0+a_1y+a_2y^2+\cdots +a_{17}y^{17})=1-(y-1)^{18}$

So $a_2$ is the $y^3$ coefficient, which, by the Binomial Theorem, is $\frac{18\cdot 17\cdot 16}{3\cdot 2\cdot 1}=3\cdot 17\cdot 16=816$

See also

1986 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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