Difference between revisions of "1986 AIME Problems/Problem 11"
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== Problem == | == 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> | + | 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 == | == Solution == | ||
− | + | === Solution 1 === | |
+ | 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]], this is <math>(-1) \cdot (-1)^{15}{18 \choose 3} = \boxed{816}</math>. | ||
+ | |||
+ | === Solution 2 === | ||
+ | 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} \\ | ||
+ | & = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}.</cmath> | ||
− | <math>y | + | 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]] tells us that this quantity is equal to <math>{18\choose 3} = \boxed{816}</math>. |
+ | === Solution 3 === | ||
+ | Again, notice <math>x=y-1</math>. Substituting <math>y-1</math> for <math>x</math> in <math>f(x)</math> gives: | ||
+ | <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> | ||
+ | From binomial theorem, the coefficient of the <math>y^2</math> term is <math>{2\choose 0} + {3\choose 1} + \cdots + {17\choose 15}</math>. This is actually the sum of the first 16 triangular numbers, which evaluates to <math>\frac{(16)(17)(18)}{6} = \boxed{816}</math>. | ||
+ | |||
+ | === Solution 4(calculus) === | ||
+ | Let <math>f(x)=1-x+x^2-x^3+\cdots+x^{16}-x^{17}</math> and <math>g(y)=a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}</math>. | ||
+ | |||
+ | Then, since <math>f(x)=g(y)</math>, | ||
+ | <cmath>\frac{d^2f}{dx^2}=\frac{d^2g}{dy^2}</cmath> | ||
+ | <math>\frac{d^2f}{dx^2} = 2\cdot 1 - 3\cdot 2x+\cdots-17\cdot 16x^{15}</math> by the power rule. | ||
+ | |||
+ | Similarly, <math>\frac{d^2g}{dy^2} = a_2(2\cdot 1) + a_3(3\cdot 2y)+\cdots+a_{17}(17\cdot 16y^{15})</math> | ||
+ | |||
+ | Now, notice that if <math>x = -1</math>, then <math>y = 0</math>, so <math>f^{''}(-1) = g^{''}(0)</math> | ||
+ | |||
+ | <math>g^{''}(0)= 2a_2</math>, and <math>f^{''}(-1) = 2\cdot 1 + 3\cdot 2 +\cdots + 16\cdot 17</math>. | ||
+ | |||
+ | Now, we can use the hockey stick theorem to see that <math>2\cdot 1 + 3\cdot 2 +\cdots + 16\cdot 17 = 2\binom{18}{3}</math> | ||
+ | |||
+ | Thus, <math>2a_2 = 2\binom{18}{3}\rightarrow a_2 = \binom{18}{3}=\boxed{816}</math> | ||
+ | |||
+ | -AOPS81619 | ||
− | |||
== See also == | == See also == | ||
− | + | {{AIME box|year=1986|num-b=10|num-a=12}} | |
− | {{ | + | [[Category:Intermediate Algebra Problems]] |
+ | {{MAA Notice}} |
Latest revision as of 01:16, 11 May 2020
Problem
The polynomial may be written in the form , where and the 's are constants. Find the value of .
Contents
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, this 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 tells us that this quantity is equal to .
Solution 3
Again, notice . Substituting for in gives: From binomial theorem, the coefficient of the term is . This is actually the sum of the first 16 triangular numbers, which evaluates to .
Solution 4(calculus)
Let and .
Then, since , by the power rule.
Similarly,
Now, notice that if , then , so
, and .
Now, we can use the hockey stick theorem to see that
Thus,
-AOPS81619
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 |
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