Difference between revisions of "2004 AMC 12B Problems/Problem 12"
(New page: {{duplicate|2004 AMC 12B #12 and 2004 AMC 10B #19}} ==Problem== In the sequence <math>2001</math>, <math>2002</math>, <math>2003</math...) |
m (→Solution: LaTeX) |
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From <math>a_2+a_3 = a_4+a_5</math> we get that <math>a_5=a_3+2</math>. | From <math>a_2+a_3 = a_4+a_5</math> we get that <math>a_5=a_3+2</math>. | ||
− | Following this pattern, we get <math>a_{2004} = a_{2002} - 2 = a_{2000} - 4 = \cdots = a_2 - 2002 = 0</math>. | + | Following this pattern, we get <math>a_{2004} = a_{2002} - 2 = a_{2000} - 4 = \cdots = a_2 - 2002 = \boxed{0}</math>. |
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== See also == | == See also == |
Revision as of 17:34, 1 January 2012
- 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 |