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)) - - 2)) +n, so = 4b+n; that is, a = 4, b = 1. Thus, we have = 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 + = 4 + n + n ) + c2n(2n). The initial conditions give 4 + Cl = 5,5 + 2Cl + 2C2 = 9, so Cl = 1, C2 = 1. Therefore, = 4 + n + 2 n + n2n = 4 + n + 2n(1 + n). 17. [BB] Try = a + bn + cn 2 We obtain a + bn + 2 = 5[a + b(n - 1) + c(n - 1)2]- 2[a + - 2) + - 2)2] + 3n 2 = 5[a - b + c + (b - 2c)n + 2 ] - - 2b + 4c + - 4c)n + 2 ] + 2 = 3a - b - 3c + (3b - + (3c + 3)n 2 . This gives the system a = - b - 3c; b = 3b - 2c; c = 3c + 3 with solution a = -3, b = -~, c = -~. Hence, a particular solution is = -3 - ~n - ~n2. The homogeneous recurrence an = 5an - 2an-2 has characteristic polynomial x 2 - 5x + 2 with characteristic roots 5±fY. Hence, = Cl(5-f11)n + C2(5±f11)n and + = ~n - ~n2 + (5-fYf + (5±fYf· The initial conditions give -3 + + = 0 and -6 + Cl(5-fY) + C2(5±fY) = 3. This gives = ~(1 - vb), = ~(1 + vb) and, hence, a = - 2n - 2n 2 + 2 (1 __ 1_) (5_v'I7)n + 2 + _1_) (5±v'I7)n n 2 2 2 v'I7 2 2 v'I7 2 . 18. We look for a particular solution of the fonnpn = a(5 n ). Substituting into the recurrence relation, we
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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.

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