Difference between revisions of "1986 AIME Problems/Problem 11"
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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> | 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> | + | Thus, <math>2a_2 = 2\binom{18}{3}\rightarrow a_2 = \binom{18}{3}=\boxed{816}</math> |
+ | |||
+ | -AOPS81619 | ||
== See also == | == See also == |
Revision as of 22:54, 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,
-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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.