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Discrete Mathematics with Graph Theory (3rd Edition) 166

# Discrete Mathematics with Graph Theory (3rd Edition) 166 -...

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164 Solutions to Exercises Cl ( -1) n + C2 (5 n ) for certain Cl, C2. Therefore, the given recurrence relation has solution - ~ (3 n ) + Cl( -l)n + c2(5 n ). The initial conditions give -~ + Cl + C2 = 4 - 2; _ Cl + 5C2 = -1 so C2 = i, Cl = 3 8 1 and the solution is an = _~(3n) + 3i (_l)n + i(5 n ). 28. (a) Note that T2 = T + 1. Thus nT 2 - nT = n. (b) Let LnT J = kl and L nT 2 J = k2 so that nT = kl + €1 and nT 2 = k2 + €2 with 0 :::; €1, €2 < 1. Then The left side of this equation is an integer, hence so is the right side. In part (a), we showed that nT 2 - nT is an integer, so €1 - €2 must be an integer too. Since, -1 < €1 - €2 < 1, it must be that €1 - €2 = 0, so LnT 2 J - LnT J = nT 2 - nT = n, as desired. 29. Dividing by an each side of the recurrence relation
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Unformatted text preview: an+l = an + an-I, we obtain a~~l = 1 + a~:l As n --+ 00, the left side has limit k while the right side has limit 1 + i. The solutions to k = 1 + i are the solutions to k 2 -k -1 = 0, namely, k = 1± 2 Y5. Since k > 0, k = 1+ 2 Y5 is the golden mean. 00 fk 30. Let S = L --k . Then 1 100 ThusS 1-----=-( 1 1) 1 100 10000 100 ' 9899 S = _1_ d S = 100 = .01010203050813213456. 10000 100 an 9899 The first ten terms of the Fibonacci sequence are clearly visible. Exercises 6.1 1. [BB] Let B be the set of connoisseurs of Canadian bacon, and A the set of those who like anchovies. (a) This asks for IB U AI = IBI + IAI - IB n AI = 10 + 7 -6 = 11. (b) This asks for IB " AI = IBI - IB n AI = 10 -6 = 4....
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