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Discrete Mathematics with Graph Theory (3rd Edition) 146

# Discrete Mathematics with Graph Theory (3rd Edition) 146 -...

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144 Solutions to Exercises 14. (a) The characteristic polynomial is x 2 + 6x + 9 with repeated characteristic root -3. Hence, an = C1 (_3)n + C2n( -3)n. SO C1 = 1, -3C1 - 3C2 = -4; that is, C1 = 1, C2 = 1/3. Thus, an = (_3)n + In( -3)n = (-3)n(1 + n/3) = -(n + 3)( _3)n-1. (b) Try Pn = a + bn + cn 2 . We obtain a + bn + cn 2 = -6[a + b(n - 1) + c(n - 1)2]- 9[a + b(n - 2) + c(n - 2)2] + n 2 + 3n = -6[a - b + c + (b - 2c)n + cn 2 ] - 9[a - 2b + 4c + (b - 4c)n + cn 2 ] + n 2 + 3n = -15a + 24b - 42c + (-15b + 48c + 3)n + (-15c + l)n 2 . Solving, we obtain a = 1 5 2 1 8' b = ~, c = 11 6 , so Pn = 1 5 2 1 8 + ~n + 1~n2. From part (a), qn = C1 (_3)n + C2n( _3)n, so Pn + qn = t2 1 8 + ~n + 116n2 + c1(-3)n + c2n(-3)n. Using the initial conditions, we obtain 51 _ 179 d 51 3 1 3 3 _ 21 128 + c1 - 128 an 128 + 8" + 16 - C1 - C2 - - 128 ' so C1 = 1, C2 = -~. Thus, an = 1 5 2 1 8 + ~n + l6 n 2 + (-3)n(1 - ~n). (c)
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