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} | + | 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> | + | |
− | \begin{align*}1 - x + x^2 + \cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \cdots - (y - 1)^{17} \\ | + | <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*}. | + | & = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}.</cmath> |
− | </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>. | 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 19:13, 28 July 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 . We want , which is the coefficient of the term in (because the in the denominator reduces the degrees in the numerator by ). 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 |