Difference between revisions of "2009 AMC 12A Problems/Problem 25"
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− | The problem asks us for the value of <math> | + | The problem asks us for the value of <math>a_{2009}</math>, and since <math>2009</math> is an odd number, we know that <math>a_{2009} = 0 \Rightarrow \boxed{\text{E}}</math>. |
== See also == | == See also == |
Revision as of 09:54, 30 July 2020
Contents
[hide]Problem
The first two terms of a sequence are and . For ,
What is ?
Solution
Consider another sequence such that , and .
The given recurrence becomes
It follows that . Since , all terms in the sequence will be a multiple of .
Now consider another sequence such that , and . The sequence satisfies .
As the number of possible consecutive two terms is finite, we know that the sequence is periodic. Write out the first few terms of the sequence until it starts to repeat.
Note that and . Thus has a period of : .
It follows that and . Thus
Our answer is .
Solution 2
(This is for the brute force users; kudos to the very intuitive solution above)
First we interpret the formula:
"The next term in the sequence is equivalent to the sum of the previous two divided by 1 minus their product"
Next, we work out the first few terms of the sequence:
So at this point, our sequence reads
Now for ..... but wait! the numerator of the next term is equal to ..... . So as long as the denominator isn't (which we can quickly verify), . Now our sequence is
Solving for :
Our sequence is now
At this point you should be getting suspicious, because some terms are repeating. Notice that the formula asks for a sum and then later a product of two terms, neither of which depend on order (commutative and associative properties). So that means if , and , then we would get the same thing if and (Basically, as long the combination of the previous two elements is the same, we should get the same result). So , , , and so on.
We can generalize this to say
"For all even such that , "
"For all odd such that , "
The problem asks us for the value of , and since is an odd number, we know that .
See also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last question |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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