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

Revision as of 20:14, 9 April 2008

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 $y=x+1$ and the $a_i$ 's are constants. Find the value of $a_2$ .

Solution

Solution 1

Using the geometric series formula, $1 - x + x^2 + \cdots - x^{17} = \frac {x^{18} - 1}{x + 1} = \frac {x^{18} - 1}{y}$ . Since $x = y - 1$ , this becomes $\frac {(y - 1)^{18} - 1}{y}$ , so we want the coefficient of the $y^3$ term in $(y - 1)^{18}$ . By the binomial theorem that is ${18 \choose 3} = \boxed{816}$ .

Solution 2

Again, notice $x = y - 1$ . So $\begin{align*}1 - x + x^2 + \cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \cdots - (y - 1)^{17} \\ & = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}.$ We want the coefficient of the $y^2$ term of each power of each binomial, which by the binomial theorem is ${2\choose 2} + {3\choose 2} + \cdots + {17\choose 2}$ . The Hockey Stick identity gives us that this quantity is equal to ${18\choose 3} = 816$ .

@@ Line 1: / Line 1: @@
 == 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>\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>.
+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 ==
-Since <math>(1+x)(1-x+x^2-x^3+\cdots +x^{16}-x^{17})=1-x^{18}</math>, we have
+=== 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(a_0+a_1y+a_2y^2+\cdots +a_{17}y^{17})=1-(y-1)^{18}</math>
+=== Solution 2 ===
+Again, notice <math>x = y - 1</math>. So
-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>
+<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 ==
 {{AIME box|year=1986|num-b=10|num-a=12}}
-* [[AIME Problems and Solutions]]
-* [[American Invitational Mathematics Examination]]
-* [[Mathematics competition resources]]
 [[Category:Intermediate Algebra Problems]]

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

Page

Toolbox

Search