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

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== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== 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>.
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Using the [[geometric series]] formula, <math>1 - x + x^2 + \cdots - x^{17} = \frac {1 - x^{18}}{1 + x} = \frac {1-x^{18}}{y}</math>. Since <math>x = y - 1</math>, this becomes <math>\frac {1-(y - 1)^{18}}{y}</math>. We want <math>a_2</math>, which is the coefficient of the <math>y^3</math> term in <math>-(y - 1)^{18}</math> (because the <math>y</math> in the denominator reduces the degrees in the numerator by <math>1</math>). By the [[binomial theorem]] that is <math>(-1) \cdot (-1)^{15}{18 \choose 3} = \boxed{816}</math>.
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=== Solution 2 ===
 
=== Solution 2 ===
 
Again, notice <math>x = y - 1</math>. So
 
Again, notice <math>x = y - 1</math>. So
<cmath>
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\begin{align*}1 - x + x^2 + \cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \cdots - (y - 1)^{17} \\
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<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*}.
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& = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}.</cmath>
</cmath>
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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>.
 
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>.
  

Revision as of 18:13, 28 July 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, $1 - x + x^2 + \cdots - x^{17} = \frac {1 - x^{18}}{1 + x} = \frac {1-x^{18}}{y}$. 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 ${18\choose 3} = 816$.

See also

1986 AIME (ProblemsAnswer KeyResources)
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
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