Discrete Mathematics University of Kentucky CS 275 Spring ... - MGNet

Now comes the second case for n = 2.
Theorem: Assume c 1 ,c 2 ,? 1 ,? 2 ,R and r 0 ,C. Suppose that r 2 4c 1 r4c 2 = 0 has one
root r 0 with multiplicity 2. Then the sequence {a n } is a solution to the recurrence
relation a n = c 1 a n-1 + c 2 a n-2 if and only if a n = ! 1
r 0
n + ! 2
nr 0
n for n,N 0 .
Example: a 0 = 1, a 1 = 6, and a n = 6a n-1 4 9a n-2 for n