Difference between revisions of "1985 AIME Problems/Problem 5"
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== Problem == | == Problem == | ||
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? | 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 == | == 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 <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 <math>a_8 = a - (a - b) = b</math>. Since the sequence is recursively defined by the first 2 terms, after this point it must continue to repeat. Thus, in particular <math>a_{j + 6} = a_j</math> for all <math>j</math>, and so repeating this <math>n</math> times, <math>a_{j + 6n} = a_j</math> for all [[integer]]s <math>n</math> and <math>j</math>. | 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 <math>a_8 = a - (a - b) = b</math>. Since the sequence is recursively defined by the first 2 terms, after this point it must continue to repeat. Thus, in particular <math>a_{j + 6} = a_j</math> for all <math>j</math>, and so repeating this <math>n</math> times, <math>a_{j + 6n} = a_j</math> for all [[integer]]s <math>n</math> and <math>j</math>. |
Revision as of 17:35, 5 June 2020
Problem
A sequence of integers is chosen so that for each . 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 and . Then , , , , and . Since the sequence is recursively defined by the first 2 terms, after this point it must continue to repeat. Thus, in particular for all , and so repeating this times, for all integers and .
Because of this, the sum of the first 1492 terms can be greatly simplified: is the largest multiple of 6 less than 1492, so , where we can make this last step because and so the entire second term of our expression is zero.
Similarly, since , .
Finally, .
Then by the givens, and so and so the answer is .
See also
1985 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |