Discrete Mathematics with Graph Theory (3rd Edition) 165

# Discrete Mathematics with Graph Theory (3rd Edition) 165 -...

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Chapter 5 Solving -4A - B = 0, A + B = 1 gives A = -~, B = ~, so 1 1( 1 ) 4( 1 ) (1 - x)(1 - 4x) = -3 1 - x + 3 1 - 4x and we obtain f(x) = -~ (_1_) (-3 + 21x) + ~ (_1_) (-3 + 21x) 3 I-x 31-4x = -Hl+x+x2+ . . ·)(-3+21x) + HI + 4x + 16x 2 + ... )( -3 + 21x) = -H -3 + 18x + ... + 18x n + ... +) + ~(-3 + 9x + 36x 2 + ... + [( -3)(4n) + 21(4n-I)]xn + ... +) = - ~ ( -3 + 18x + ... + 18x n + ... + ) + H -3 + 9x + 36x 2 + ... + 9(4n-l)xn + ... +) = -3 + 6x + 42x2 + ... + (-6 + 3(4n))xn + ... + so we obtain an = -6 + 3(4n) as before. 163 25. First we find a particular solution Pn. Since f(n) = 5 is linear, we try a linear function Pn = a + bn for Pn. Substituting in the recurrence relation, we have a + bn = 3[a + b(n - 1)]- 2[a + b(n - 2)] + 5 = a + bn + b + 5, for which b = -5 and any a is a solution. We take a = 0 and Pn = -5n. The corresponding homogeneous recurrence relation is an = 3an -1 - 2an-2, whose characteristic polynomial is x 2 - 3x + 2 = (x -1)(x
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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)n-l. 27. We first try Pn = a(3 n ). Substituting gives a(3 n ) = 4a(3 n-l ) + 5a(3 n-2 ) + 3 n 9a = 12a + 5a + 9, so a = -~ and Pn = -~(3n) is a particular solution. Next we solve an = 4an-1 + 5an-2. The characteristic polynomial, x 2 -4x -5, has distinct roots Xl = -1, X2 = 5, so the solution is qn =...
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