Difference between revisions of "2004 AMC 12B Problems/Problem 12"
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We already know that <math>a_1=2001</math>, <math>a_2=2002</math>, <math>a_3=2003</math>, and <math>a_4=2000</math>. Let's compute the next few terms to get the idea how the sequence behaves. We get <math>a_5 = 2002+2003-2000 = 2005</math>, <math>a_6=2003+2000-2005=1998</math>, <math>a_7=2000+2005-1998=2007</math>, and so on. | We already know that <math>a_1=2001</math>, <math>a_2=2002</math>, <math>a_3=2003</math>, and <math>a_4=2000</math>. Let's compute the next few terms to get the idea how the sequence behaves. We get <math>a_5 = 2002+2003-2000 = 2005</math>, <math>a_6=2003+2000-2005=1998</math>, <math>a_7=2000+2005-1998=2007</math>, and so on. | ||
− | We can now discover the following pattern: <math>a_{2k+1} = | + | We can now discover the following pattern: <math>a_{2k+1} = 2001+2k</math> and <math>a_{2k}=2004-2k</math>. This is easily proved by induction. It follows that <math>a_{2004}=a_{2\cdot 1002} = 2004 - 2\cdot 1002 = \boxed{0}</math>. |
=== Solution 2 === | === Solution 2 === |
Revision as of 21:35, 25 May 2020
- 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
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 .
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.