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520  8 / Advanced Counting Techniques

                                 EXAMPLE 7      Suppose that the roots of the characteristic equation of a linear homogeneous recurrence relation
                                                are 2, 2, 2, 5, 5, and 9 (that is, there are three roots, the root 2 with multiplicity three, the root
                                                5 with multiplicity two, and the root 9 with multiplicity one). What is the form of the general
                                                solution?

                                                Solution: By Theorem 4, the general form of the solution is

                                                                         n
                                                                      2
                                                                                                  n
                                                                                          n
                                                    (α 1,0 + α 1,1 n + α 1,2 n )2 + (α 2,0 + α 2,1 n)5 + α 3,0 9 .
                                                                                                                               ▲
                                                    We now illustrate the use of Theorem 4 to solve a linear homogeneous recurrence relation
                                                with constant coefficients when the characteristic equation has a root of multiplicity three.
                                 EXAMPLE 8      Find the solution to the recurrence relation


                                                    a n =−3a n−1 − 3a n−2 − a n−3
                                                with initial conditions a 0 = 1, a 1 =−2, and a 2 =−1.

                                                Solution: The characteristic equation of this recurrence relation is

                                                          2
                                                     3
                                                    r + 3r + 3r + 1 = 0.
                                                              2
                                                                                3
                                                         3
                                                Because r + 3r + 3r + 1 = (r + 1) , there is a single root r =−1 of multiplicity three of
                                                the characteristic equation. By Theorem 4 the solutions of this recurrence relation are of the
                                                form
                                                                                     2
                                                                                          n
                                                                             n
                                                                n
                                                    a n = α 1,0 (−1) + α 1,1 n(−1) + α 1,2 n (−1) .
                                                To find the constants α 1,0 ,α 1,1 , and α 1,2 , use the initial conditions. This gives
                                                    a 0 = 1 = α 1,0 ,
                                                    a 1 =−2 =−α 1,0 − α 1,1 − α 1,2 ,
                                                    a 2 =−1 = α 1,0 + 2α 1,1 + 4α 1,2 .

                                                The simultaneous solution of these three equations is α 1,0 = 1, α 1,1 = 3, and α 1,2 =−2.
                                                Hence, the unique solution to this recurrence relation and the given initial conditions is the
                                                sequence {a n } with

                                                                          n
                                                                    2
                                                    a n = (1 + 3n − 2n )(−1) .
                                                                                                                               ▲
                                                Linear Nonhomogeneous Recurrence Relations
                                                with Constant Coefficients

                                                We have seen how to solve linear homogeneous recurrence relations with constant coefficients.
                                                Is there a relatively simple technique for solving a linear, but not homogeneous, recurrence
                                                relation with constant coefficients, such as a n = 3a n−1 + 2n? We will see that the answer is yes
                                                for certain families of such recurrence relations.
                                                    The recurrence relation a n = 3a n−1 + 2n is an example of a linear nonhomogeneous
                                                recurrence relation with constant coefficients, that is, a recurrence relation of the form


                                                    a n = c 1 a n−1 + c 2 a n−2 + ··· + c k a n−k + F(n),
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