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Discrete Mathematics with Graph Theory (3rd Edition) 120

# Discrete Mathematics with Graph Theory (3rd Edition) 120 -...

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118 Solutions to Exercises that 1(3)(5)··· (2k - 1) ;::: 2(4)(6)··· (2k - 2). We have 1(3)(5) ... [2(k + 1) - 1] = 1(3)(5) ... (2k + 1) = (1(3)(5) ... (2k - 1))(2k + 1) ;::: (2(4)(6) . . · (2k - 2))(2k + 1) using the induction hypothesis, ;::: 2(4)(6)··· (2k - 2)(2k), which is the desired inequality with n = k + 1 (since 2k = 2(k + 1) - 2). By the Principle of Mathematical Induction, 1(3)(5)··· (2n - 1) ;::: 2(4)(6)··· (2n - 2) for every integer n ;::: 2. 1 10. (a) [BB] For n = 1, I)Xi + Yi) is, by convention, just the single term Xl + Y1. Since this equals i=l 1 1 LXI + L YI the formula holds for n = 1. Now suppose that k ;::: 1 and the formula holds for i=l i=l n = k. Then k+1 k L(Xi + Yi) = [L(Xi + Yi)] + (Xk+1 + Yk+1) i=l i=l k k = LXi + LYi + (Xk+1 +
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Unformatted text preview: i=l k k k+1 k+1 = [(LXi) +Xk+1] + [(LYi) +Yk+1] = LXi + LYi, i=l i=l i=l i=l which is the formula with n = k + 1. Thus, the result holds for all n ;::: 1 by the Principle of Mathematical Induction. (b) For n = 1, 2:~=1 CXi = 2::=1 CXi = CX1 = C 2::=1 Xi and the formula is true. Now suppose that k ;::: 1 and the formula is true for n = k. Then k+1 k k L CXi = (L CXi) + CXk+1 = (c L Xi) + CXk+1, (by the induction hypothesis) i=l i=l i=l k k+1 =C[(LXi) +Xk+1] =CLXi, i=l i=l which is the formula with n = k + 1. Thus, the result holds for all n ;::: 1 by the Principle of Mathematical Induction. n 2 (c) For n = 2, L(Xi -Xi-I) = L(Xi -Xi-I) which is, by convention, the single term X2 -X2-1 = i=2 i=2 X2 -Xl = Xn -Xl. Thus, the formula is true for n = 2. Now suppose that k ;::: 2 and the formula...
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