2004 AMC 12B Problems/Problem 12
- The following problem is from both the 2004 AMC 12B #12 and 2004 AMC 10B #19, so both problems redirect to this page.
Problem
In the sequence , , , , each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is . What is the term in this sequence?
Solution 1
We already know that , , , and . Let's compute the next few terms to get the idea how the sequence behaves. We get , , , and so on.
We can now discover the following pattern: and . This is easily proved by induction. It follows that .
Solution 2
Note that the recurrence can be rewritten as .
Hence we get that and also
From the values given in the problem statement we see that .
From we get that .
From we get that .
Following this pattern, we get .
Solution 3
Our recurrence is , so we get , so , so our formula for the recurrence is .
Substituting our starting values gives us .
So,
See also
2004 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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 |
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.