Difference between revisions of "2004 AMC 12B Problems/Problem 12"
(→Solution 3) |
(→Solution 1) |
||
Line 13: | Line 13: | ||
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>. | 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 1 but in a more not smart way== | ||
+ | |||
+ | Subtract 2000 from each of the terms so the sequence turns into 1, 2, 3, 0, 5, -2, 7, -4, 9, -6... (1+2-3=0, 2+3-0=5, 3+0-5=-2, 0+5-(-2)=7, 5+(-2)-7=-4, etc.). Quickly notice that after taking away 1 and 2, the 2nd term is 0, the 4th term is -2, the 6th term is -4, the 8th term is -6, etc. Thus, in the sequence without 1 and 2, the (2n)th term has a value of -2(n-1). Therefore, to find the 2004th term in the original sequence, we take away the first 2 values to form the new sequence so the value of the 2004th term in the original sequence is the 2002nd term in the new sequence. 2002 is 2 times 1001. Thus, the value is -2(1001-1)= -2000. BUT, REMEMBER THAT AT THE START WE TOOK AWAY 2000 FROM EACH TERM. ADD -2000 TO 2000 TO GET <math>\boxed{0}</math>. | ||
== Solution 2 == | == Solution 2 == |
Latest revision as of 10:49, 16 August 2024
- The following problem is from both the 2004 AMC 12B #12 and 2004 AMC 10B #19, so both problems redirect to this page.
Contents
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 1 but in a more not smart way
Subtract 2000 from each of the terms so the sequence turns into 1, 2, 3, 0, 5, -2, 7, -4, 9, -6... (1+2-3=0, 2+3-0=5, 3+0-5=-2, 0+5-(-2)=7, 5+(-2)-7=-4, etc.). Quickly notice that after taking away 1 and 2, the 2nd term is 0, the 4th term is -2, the 6th term is -4, the 8th term is -6, etc. Thus, in the sequence without 1 and 2, the (2n)th term has a value of -2(n-1). Therefore, to find the 2004th term in the original sequence, we take away the first 2 values to form the new sequence so the value of the 2004th term in the original sequence is the 2002nd term in the new sequence. 2002 is 2 times 1001. Thus, the value is -2(1001-1)= -2000. BUT, REMEMBER THAT AT THE START WE TOOK AWAY 2000 FROM EACH TERM. ADD -2000 TO 2000 TO GET .
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.