Difference between revisions of "1985 AIME Problems/Problem 5"

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== Problem ==
 
== Problem ==
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A [[sequence]] of [[integer]]s <math>a_1, a_2, a_3, \ldots</math> is chosen so that <math>a_n = a_{n - 1} - a_{n - 2}</math> for each <math>n \ge 3</math>. What is the sum of the first 2001 terms of this sequence if the sum of the first 1492 terms is 1985, and the sum of the first 1985 terms is 1492?
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== Solution ==
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The problem gives us a sequence defined by a [[recursion]], so let's calculate a few values to get a feel for how it acts.  We aren't given initial values, so let <math>a_1 = a</math> and <math>a_2 = b</math>.  Then <math>a_3 = b - a</math>, <math>a_4 = (b - a) - b = -a</math>, <math>a_5 = -a - (b - a) = -b</math>, <math>a_6 = -b - (-a) = a - b</math>, <math>a_7 = (a - b) - (-b) = a</math> and ...
  
== Solution ==
 
 
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{{solution}}
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== See also ==
 
== See also ==
 
* [[1985 AIME Problems/Problem 4 | Previous problem]]
 
* [[1985 AIME Problems/Problem 4 | Previous problem]]
 
* [[1985 AIME Problems/Problem 6 | Next problem]]
 
* [[1985 AIME Problems/Problem 6 | Next problem]]
 
* [[1985 AIME Problems]]
 
* [[1985 AIME Problems]]
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[[Category:Intermediate Algebra Problems]]

Revision as of 12:56, 18 November 2006

Problem

A sequence of integers $a_1, a_2, a_3, \ldots$ is chosen so that $a_n = a_{n - 1} - a_{n - 2}$ for each $n \ge 3$. What is the sum of the first 2001 terms of this sequence if the sum of the first 1492 terms is 1985, and the sum of the first 1985 terms is 1492?

Solution

The problem gives us a sequence defined by a recursion, so let's calculate a few values to get a feel for how it acts. We aren't given initial values, so let $a_1 = a$ and $a_2 = b$. Then $a_3 = b - a$, $a_4 = (b - a) - b = -a$, $a_5 = -a - (b - a) = -b$, $a_6 = -b - (-a) = a - b$, $a_7 = (a - b) - (-b) = a$ and ...

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See also