formula for sequences satisfying second order recurrence relations
Theorem.
Let be a sequence such that there exist constants with and, for all positive integers ,
Then for all nonnegative integers , we have
where denotes the floor function and denotes the binomial coefficient.
Proof.
This will be proven by induction. Due to the presence of , odd and even should be considered separately. Thus, there are two base steps:
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If , then
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If , then
Now assume that the theorem holds for all nonnegative integers less than or equal to some positive integer . We must show that the theorem holds for .
Recall that
We can use the induction hypothesis on and :
First assume that is odd. Then the first of the two summations has terms and the second summation has terms. Therefore, we must split off the term from the second summation before combining the two summations:
Now assume that is even. Then the two summations have the same number of terms and can be combined as is:
Hence, the theorem holds for all nonnegative integers . ∎
Title | formula for sequences satisfying second order recurrence relations |
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Canonical name | FormulaForSequencesSatisfyingSecondOrderRecurrenceRelations |
Date of creation | 2013-03-22 17:51:43 |
Last modified on | 2013-03-22 17:51:43 |
Owner | Wkbj79 (1863) |
Last modified by | Wkbj79 (1863) |
Numerical id | 7 |
Author | Wkbj79 (1863) |
Entry type | Theorem |
Classification | msc 11B37 |
Classification | msc 03D20 |