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

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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>.
 
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>.
 
== Solution ==
 
== Solution ==
{{solution}}
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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>
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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 ==
 
* [[1986 AIME Problems]]
 
* [[1986 AIME Problems]]
  
 
{{AIME box|year=1986|num-b=10|num-a=12}}
 
{{AIME box|year=1986|num-b=10|num-a=12}}

Revision as of 20:52, 26 March 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 thet $\displaystyle a_i$'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 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AIME Problems and Solutions
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