Difference between revisions of "1990 AIME Problems/Problem 15"
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− | + | A [[linear recurrence | recurrence]] of the form <math>T_n=AT_{n-1}+BT_{n-2}</math> will have the closed form <math>T_n=ax^n+by^n</math>, where <math>x,y</math> are the values of the starting term that make the sequence geometric, and <math>a,b</math> are the appropriately chosen constants such that those special starting terms linearly combine to form the actual starting terms. | |
Suppose we have such a recurrence with <math>T_1=3</math> and <math>T_2=7</math>. Then <math>T_3=ax^3+by^3=16=7A+3B</math>, and <math>T_4=ax^4+by^4=42=16A+7B</math>. | Suppose we have such a recurrence with <math>T_1=3</math> and <math>T_2=7</math>. Then <math>T_3=ax^3+by^3=16=7A+3B</math>, and <math>T_4=ax^4+by^4=42=16A+7B</math>. |
Revision as of 22:34, 2 June 2011
Contents
Problem
Find if the real numbers , , , and satisfy the equations
Solution 1
Set and . Then the relationship
can be exploited:
Therefore:
Consequently, and . Finally:
Solution 2
A recurrence of the form will have the closed form , where are the values of the starting term that make the sequence geometric, and are the appropriately chosen constants such that those special starting terms linearly combine to form the actual starting terms.
Suppose we have such a recurrence with and . Then , and .
Solving these simultaneous equations for and , we see that and . So, .
See also
1990 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |