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Discrete Mathematics with Graph Theory (3rd Edition) 147

Discrete Mathematics with Graph Theory (3rd Edition) 147 -...

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Section 5.3 145 16. We try Pn = a +bn and obtain a+ bn = 4(a+ b(n -1)) - 4(a+ b(n - 2)) +n, so a+ bn = 4b+n; that is, a = 4, b = 1. Thus, we have Pn = 4 + n as a particular solution. As seen in Problem 21, the homogeneous recurrence an = 4an -l - 4an -2 has solution qn = cl(2 n ) + c2n(2n), so Pn + qn = 4 + n + cl(2 n ) + c2n(2n). The initial conditions give 4 + Cl = 5,5 + 2Cl + 2C2 = 9, so Cl = 1, C2 = 1. Therefore, an = 4 + n + 2 n + n2n = 4 + n + 2n(1 + n). 17. [BB] Try Pn = a + bn + cn 2 We obtain a + bn + cn 2 = 5[a + b(n - 1) + c(n - 1)2]- 2[a + b(n - 2) + c(n - 2)2] + 3n 2 = 5[a - b + c + (b - 2c)n + cn 2 ] - 2[a - 2b + 4c + (b - 4c)n + cn 2 ] + 3n 2 = 3a - b - 3c + (3b - 2c)n + (3c + 3)n 2 . This gives the system a = 3a - b - 3c; b = 3b - 2c; c = 3c + 3 with solution a = -3, b = -~, c = -~. Hence, a particular solution is Pn = -3 - ~n - ~n2. The homogeneous recurrence an = 5an -l - 2an -2 has characteristic polynomial x 2 - 5x + 2 with characteristic roots 5±fY. Hence, qn = Cl(5-f11)n + C2(5±f11)n and Pn + qn = -3 - ~n - ~n2 + Cl (5-fYf + C2 (5±fYf· The initial conditions give -3 + Cl + C2 = 0 and -6 + Cl(5-fY) + C2(5±fY) = 3. This gives Cl = ~(1 - vb), C2 = ~(1 + vb) and, hence, a = -3 - 2n - 2n 2 + 2 (1 __ 1_)
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