Section 5.4 (b) We have as desired. 2x[c(n -1)xn-1J-x2[c(n -2)xn-2J 2c(n -l)xn -c(n -2)xn = cnxn = an 149 (c) By part (b), C2nXn is one solution and by part (a) of Exercise 25 CIXn is another. Hence, by part (b) of Exercise 25, we conclude that CIXn + c2nxn is also a solution. Finally, two initial conditions will detennine Cl and C2 because they detennine two linear equations in the two unknowns Cl, C2 which are guaranteed to have a unique solution. 27. (a) We have Pn = rPn-l + SPn-2 + f(n) for n :2: 2 and also qn = rqn-l + sqn-2 for n :2: 2. Hence, rPn-l + SPn-2 + f(n) + rqn-l + sqn-2 r(Pn-l + qn-l) + S(Pn-2 + qn-2) + f(n) which says that Pn + qn satisfies the given relation for n :2: 2. (b) Part (a) shows that Pn + qn satisfies the given recurrence relation where qn = CIXl' + C2X~ or qn = CIXn + C2nXn according as the polynomial x2 -rx -s has two distinct roots or just one. Two initial conditions provide two equations which can be solved for Cl, C2. Exercises 5.4 1. (a) [BB] Since (2 -3x)2 = 4 -12x + 9x2 the associated sequence is
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