Difference between revisions of "1986 AIME Problems/Problem 11"
Scorpius119 (talk | contribs) (solution added) |
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So <math>a_2</math> is the <math>y^3</math> coefficient, which, by the Binomial Theorem, is <math>\frac{18\cdot 17\cdot 16}{3\cdot 2\cdot 1}=3\cdot 17\cdot 16=816</math> | So <math>a_2</math> is the <math>y^3</math> coefficient, which, by the Binomial Theorem, is <math>\frac{18\cdot 17\cdot 16}{3\cdot 2\cdot 1}=3\cdot 17\cdot 16=816</math> | ||
== See also == | == See also == | ||
| − | |||
| − | |||
{{AIME box|year=1986|num-b=10|num-a=12}} | {{AIME box|year=1986|num-b=10|num-a=12}} | ||
| + | * [[AIME Problems and Solutions]] | ||
| + | * [[American Invitational Mathematics Examination]] | ||
| + | * [[Mathematics competition resources]] | ||
Revision as of 13:52, 6 May 2007
Problem
The polynomial 
may be written in the form 
, where 
and thet 
's are constants. Find the value of 
.
Solution
Since 
, we have
So 
is the 
coefficient, which, by the Binomial Theorem, is 
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 | ||
