1986 AIME Problems/Problem 11

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 ${18\choose 3} = 816$ .

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
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1986 AIME Problems/Problem 11

Problem

Contents

Solution

Solution 1

Solution 2

See also