Unformatted text preview:  2), with roots Xl = 1 and X2 = 2. The solution to the homogeneous recurrence relation is qn = CI (1 n) + c2(2 n ) = CI + C22n. Thus, the solution to the given recurrence is The initial conditions give 1 = CI + C2 and 1 = CI + 2C2 5, so CI 4 + 5(2n) 5n. 26. (a) The first five terms are 2, I, 12, 81, 378. 5 and an (b) The characteristic polynomial x 2 + 6x + 9 = (x + 3)2 has a repeated root, so the solution takes the form an = cI(3)n+c2n(3)n. The initial conditions give ao = 2 = CI andal = 1 = 3CI 3C2, so 3C2 = 1  3CI = 5 and C2 = i. The solution is an = (2)3 n + in( _3)n = (2)3 n 5n( _3)nl. 27. We first try Pn = a(3 n ). Substituting gives a(3 n ) = 4a(3 nl ) + 5a(3 n2 ) + 3 n 9a = 12a + 5a + 9, so a = ~ and Pn = ~(3n) is a particular solution. Next we solve an = 4an1 + 5an2. The characteristic polynomial, x 2 4x 5, has distinct roots Xl = 1, X2 = 5, so the solution is qn =...
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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