Discrete Mathematics with Graph Theory (3rd Edition) 114

Discrete Mathematics with Graph Theory (3rd Edition) 114 -...

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112 Solutions to Exercises (e) When n = 1, 1 1 1 1 1 1 1· 2 + 2·3 + 3·4 + . .. + n(n + 1) = 1· 2 = 2 and so the result holds. Suppose the result is true for n = k ;::: 1; that is, suppose 1 1 1 1 k 1·2 + 2·3 + 3·4 + . .. + k(k + 1) = k + We must prove that the result is true for n = k + 1, that is, that 1 1 1 1 k+l 1·2 + 2·3 + 3·4 + . .. + (k + 1)[(k + 1) + 1] = (k + 1) + Now 1 1 1 1 1·2 + 2·3 + 3·4 + . .. + (k + 1)[(k + 1) + 1] [ 1 1 1 1] 1 = 1·2 + 2·3 + 3·4 + . .. + k(k + 1) + (k + 1)[(k + 1) + 1] = _k_ + (k )~k ) (by the induction hypothesis) k+l +1 +2 k(k+2)+1 (k+l)2 k+l (k+l)(k+2) (k+l)(k+2) k+2 as desired. By the Principle of Mathematical Induction, the assertion is true for all n ;::: 1. (1) [BB] For n = 1, the left side is ~ = 1 - -fr' so the formula is correct. Now assume k ;::: 1 and that the formula holds for n = k; that is, assume that
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Unformatted text preview: 2 2 2 2 1 3" + "9 + 27 + . .. + 3k = 1 -3k With n = k + 1, the left side of the sum in question is 2 2 2 2 1 2 3" +"9 + 27 + . .. + 3k+l = (1-3k ) + 3k+ 1 3k+l -1 1 3k+ 1 = 1 -3k+ 1 ' which is the desired formula. By the Principle of Mathematical Induction, the formula holds for all n ;::: 1. (g) [BB] When n = 1, the left side is 6 - 2 = 4 and the right side is 1(4), so the formula holds. Now assume that k ;::: 1 and that 4 + 10 + 16 + . .. + (6k -2) = k(3k + 1). We have 4 + 10 + 16 + . .. + [6(k + 1) - 2] = 4 + 10 + 16 + . .. + (6k + 4) = (4 + 10 + 16 + . .. + (6k -2)) + (6k + 4) = k(3k + 1) + (6k + 4) using the induction hypothesis = 3k 2 + 7k + 4 = (k + 1)(3k + 4) which is the given formula with n = k + 1. By the Principle of Mathematical Induction, we conclude that 4 + 10 + 16 + . .. + (6n - 2) = n(3n + 1) for all n ;::: 1....
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