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

Revision as of 19:05, 4 July 2013

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, $1 - x + x^2 + \cdots - x^{17} = \frac {1 - x^{18}}{1 + x} = \frac {1-x^{18}}{y}$ . Since $x = y - 1$ , this becomes $\frac {1-(y - 1)^{18}}{y}$ . We want $a_2$ , which is the coefficient of the $y^3$ term in $-(y - 1)^{18}$ (because the $y$ in the denominator reduces the degrees in the numerator by $1$ ). By the binomial theorem that is $(-1) \cdot (-1)^{15}{18 \choose 3} = \boxed{816}$ .

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 ${18\choose 3} = \boxed{816}$ .

Revision as of 20:20, 11 April 2012 (view source) Baijiangchen (talk \| contribs) m (→‎Solution 2) ← Older edit		Revision as of 19:05, 4 July 2013 (view source) Nathan wailes (talk \| contribs) Newer edit →
Line 19:		Line 19:

	[[Category:Intermediate Algebra Problems]]		[[Category:Intermediate Algebra Problems]]
		+	{{MAA Notice}}

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