Discrete Mathematics with Graph Theory (3rd Edition) 112

Discrete Mathematics with Graph Theory (3rd Edition) 112 -...

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110 Then Solutions to Exercises 1 3 + 2 3 + . .. + (k + 1)3 = (13 + 2 3 + . .. + k 3 ) + (k + 1)3 = k 2 (k + 1)2 + (k + 1)3 (by the induction hypothesis) 4 k 2 (k + 1)2 + 4(k + 1)3 4 (k + 1)2(k 2 + 4(k + 1) 4 which is the given statement with n = k + 1. So by the Principle of Mathematical Induction, we n 2 (n + 1)2 conclude that 1 3 + 2 3 + . .. + n 3 = 4 for all n 2: 1. ( n(n+1»)2 n 2 (n+1)2 (c) This follows directly from (a) and (b) and the fact that -2- = 4 . 6. (a) [BB] If n = 1, 1 + 21 + 22 + 2 3 + . .. + 2 n = 1 + 2 = 3, while 21+1 - 1 = 4 - 1 = 3, so the result is true if n = 1. Suppose the result is true for n = k 2: 1; that is, suppose 1 + 2 + 22 + 2 3 + . .. + 2k = 2k+1 -1. We wish to prove the result for n = k + 1; that is, we wish to prove that 1 + 2 + 22 + 2 3 + . .. + 2k+1 = 2(k+1)+1 - 1. Now 1 + 2 + 22 + 2 3 + . .. + 2k+1 = (1 + 2 + 22 + 2 3 + . .. + 2k) + 2k+1 = (2k+1 - 1) + 2k+1 (by the induction hypothesis) = 2 . 2k+1 - 1 = 2k+2 - 1 = 2(k+1)+1
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Unformatted text preview: -1 as required. By the Principle of Mathematical Induction, we conclude that the given assertion is true for all n 2: 1. (b) [BB] If n = 1,1 2 -22 + 3 2 -4 2 + . .. + (_I)n-1 n2 = 12 = 1, by definition, while (_I)n-1 n(n 2 +1) = (_1)1-11(1i 1 ) = 1, and so the result is true for n = 1. Now suppose that k 2: 1 and the result is true for n = k; that is, suppose that 12 _ 22 + 3 2 _ 42 + . .. + (_I)k-1k2 = (-ll-1 k(k + 1). 2 We wish to prove that 12 -22 + 3 2 -4 2 + . .. + (_I)(k+1)-1(k + 1)2 = (_1)(k+1)-1 (k + 1)[(k + 1) + 1] 2 . Now 12 -22 + 3 2 -4 2 + . .. + (_I)(k+1)-1(k + 1)2 = [12 _ 22 + 3 2 _ 4 2 + . .. + (_I)k-1k2] + (_I)(k+1)-1(k + 1)2 = (_I)k-1 k(k + 1) + (-I)k(k + 1)2 2 (using the induction hypothesis) = (_I)k-1 k 2 + k + (_I)k-1( -1)(k 2 + 2k + 1) 2 = (_I)k_1(k 2 ;k -k 2-2k-l) = (_I)k-1 k 2 + k -2k2 -4k - 2 2 = (_I)k-1 (-k 2 -2 3k - 2) = (_I)k (k + 1)2(k + 2)...
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