Discrete Mathematics with Graph Theory (3rd Edition) 128

Discrete Mathematics with Graph Theory (3rd Edition) 128 -...

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126 Solutions to Exercises 35. [BB] (using the result of Exercise 34) n IT (1 + (±1)2r) = 2n. r=1 IT n 2r - 1 3 5 2n - 1 36. -- = - . - ... -- = 2n - 1. r=2 2r - 3 1 3 2n - 3 Proof. If n = 2, Il~=2 ~;=~ = f = 3 = 2(2) - 1 and so the result is true. Now suppose that k ::::: 1 and the result is true for n = k; that is, suppose Il~=2 ~;=~ = 2k - 1. We must prove it is true for n = k + 1; that is, we must prove Il~~~ ~;=~ = 2(k + 1) - 1 = 2k + 1. We have k+1 k IT 2r-1 = (IT 2r-1)(2(k+1)-1) =(2k_1)(2k+1) =2k+1 r=2 2r - 3 r=2 2r - 3 2(k + 1) - 3 2k - 1 as claimed. By the Principle of Mathematical Induction, the result is true for all n ::::: 2. n-I 37. (a) For n = 1, IT Fr = Fo = 3 = 5 - 2 = FI - 2, so the formula is correct. r=O k-I Now let k ::::: 1 and assume that IT Fr = Fk - 2. Then r=O k k-I IT Fr = Fk IT Fr = Fk(Fk - 2) (by the induction hypothesis) r=O r=O = Ff - 2Fk = (22k + 1)2 - 2(22k + 1) = 22 k + 1 + 2(22k) + 1 _ 2(22k) _ 2 = 22 k + 1 + 1 - 2 = F k + 1 -
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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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