Difference between revisions of "1986 AIME Problems/Problem 11"
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== Solution 4(calculus) == | == Solution 4(calculus) == | ||
− | Let f(x) | + | 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}=</math> | ||
== See also == | == See also == |
Revision as of 23:53, 3 March 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,
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.