Difference between revisions of "1986 AIME Problems/Problem 11"
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== Problem == | == Problem == | ||
− | 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> | + | 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>y=x+1</math> and the <math>a_i</math>'s are [[constant]]s. Find the value of <math>a_2</math>. |
+ | __TOC__ | ||
== Solution == | == Solution == | ||
− | + | === Solution 1 === | |
− | + | Using the geometric series formula, <math>1 - x + x^2 + \cdots - x^{17} = \frac {x^{18} - 1}{x + 1} = \frac {x^{18} - 1}{y}</math>. Since <math>x = y - 1</math>, this becomes <math>\frac {(y - 1)^{18} - 1}{y}</math>, so we want the coefficient of the <math>y^3</math> term in <math>(y - 1)^{18}</math>. By the [[binomial theorem]] that is <math>{18 \choose 3} = \boxed{816}</math>. | |
− | <math>y | + | === Solution 2 === |
− | + | Again, notice <math>x = y - 1</math>. So | |
− | + | <cmath> | |
+ | \begin{align*}1 - x + x^2 + \cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \cdots - (y - 1)^{17} \\ | ||
+ | & = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}. | ||
+ | </cmath> | ||
+ | We want the coefficient of the <math>y^2</math> term of each power of each binomial, which by the binomial theorem is <math>{2\choose 2} + {3\choose 2} + \cdots + {17\choose 2}</math>. The [[Hockey Stick identity]] gives us that this quantity is equal to <math>{18\choose 3} = 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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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] |
Revision as of 21:14, 9 April 2008
Problem
The polynomial may be written in the form , where and the 's are constants. Find the value of .
Solution
Solution 1
Using the geometric series formula, . Since , this becomes , so we want the coefficient of the term in . By the binomial theorem that is .
Solution 2
Again, notice . So We want the coefficient of the term of each power of each binomial, which by the binomial theorem is . The Hockey Stick identity gives us that this quantity is equal to .
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 |