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Discrete Mathematics with Graph Theory (3rd Edition) 145

# Discrete Mathematics with Graph Theory (3rd Edition) 145 -...

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Section 5.3 143 (b) [BB] The function f(n) = 24 is constant, so we try Pn = k. This gives k = -2k + 15k + 24, so k = -2 = Pn. In part (a), we saw that the corresponding homogeneous recurrence relation has solution qn = c1(3 n ) + C2( -5)n, so an = Pn + qn = c1(3 n ) + C2( _5)n - 2. The initial conditions give 1 = ao = C1 + C2 - 2 and -1 = a1 = 3C1 - 5C2 - 2, so C1 = 2, C2 = 1 and our solution is an = 2(3 n ) + (_5)n - 2. 11. (a) To fit this problem to Theorem 5.3.1, we rewrite the given recurrence as an = -8a n -1 -16a n -2, n ?: 2. The characteristic polynomial is x 2 + 8x + 16 with repeated characteristic root -4. We obtain an = C1 (_4)n + C2n( -4)n. Since C1 = 5, -4C1 - 4C2 = 17, we get C1 = 5, C2 = -37/4. The solution is an = 5( _4)n - 31 n( _4)n = (_4)n 20- 4 37n = (_4)n-1 (37n - 20). (b) Since f(n) = 5 is constant, we try Pn = k and obtain k = -8k - 16k + 5, so k = g = Pn. From part (a), we know that the corresponding homogeneous recurrence relation has solution qn = C1 (_4)n + C2n( _4)n so an = Pn + qn = C1 (_4)n + C2n( -4)n + g. The initial conditions give 2 = ao = C1 + g and -1 = a1 = -4C1 - 4C2 + g, so C1 = ~ and C2 = -lOur solution is an = ~(-4)n - ~n( -4)n + g.
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