Discrete Mathematics with Graph Theory (3rd Edition) 116

Discrete Mathematics with Graph Theory (3rd Edition) 116 -...

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114 Solutions to Exercises k+1 k Now L(i + 1)2i = + 1)2i + (k + 2)2k+1 i=1 = k2k+1 + + 2)2k+1(using the induction hypothesis) = (k + k + = (2k + = 2(k + 1)2k+1 = + 1)2k+2 as required. By the Principle of Mathematical Induction, the given assertion is true for all n 2: 1. (b) When n = 1, n i2 12 1 1 8 (2i -1)(2i + 1) = [2(1) -1][2(1) + 1] = 1(3) = 3' while ~~';:,~V) = ~ = i and so the result holds. Now suppose that k 2: 1 and the result is true for n = k; thus, we suppose that k k(k+1) 8 (2i - 1)(2i + 1) = 2(2k + 1) . We have to show that the result holds for n = k + 1, that is, that i 2 + l)[(k + + 8 (2i - 1)(2i
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