Problem 27 · 2010 Math Kangaroo
Stretch
Algebra & Patterns
arithmetic-sequencesubstitution
In a sequence the first three terms are 1, 2 and 3. From the fourth term onwards each term is found from the three previous terms: the third one back is subtracted from the sum of the two before it. This gives the sequence 1, 2, 3, 0, 5, −2, 7, … What is the 2010th term of this sequence?
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Answer: A — −2006
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Hint 1 of 2
Compute a few more terms with the rule (sum of the two before, minus the one three back).
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Hint 2 of 2
Watch how odd- and even-numbered terms behave separately.
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Approach: find the pattern by position
- The rule a(n) = a(n-3) + a(n-2) - a(n-1) gives 1,2,3,0,5,-2,7,-4,9,-6,...
- Odd places give a(n) = n; even places give a(n) = 4 - n.
- Term 2010 is even: a(2010) = 4 - 2010 = -2006.
Mark:
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