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 == | ||
− | {{ | + | 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 may be written in the form , where and thet 's are constants. Find the value of .
Solution
Since , we have
So is the coefficient, which, by the Binomial Theorem, is
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 |