Unformatted text preview: n = k + 1; that is, we must prove that (1 + x)k+1 2: 1 + (k + l)x. Now (1 + x)k+1 (1 + x)k(l + x) > (1 + kx)(l + x) (by the induction hypothesis) 1 + kx + x + kx 2 > 1 + (k + l)x since kx 2 2: 0, and this is what we wanted. So, by the Principle of Mathematical Induction, the given statement is true for all n 2: 1. (f) For n = 2, 1 1 1 1 1 7 13 + + . .. + = +  = > . n + 1 n + 2 2n 3 4 12 24 Now suppose k 2: 2 and the inequality holds for k; that is, suppose that We must prove that the inequality holds for k + 1; that is, we must prove that 1 1 1 1 13 k + 2 + k + 3 + k + 4 + ... + 2k + 2 > 24· Now, 1 1 1 1 1 1 1 = k + 1 + k + 2 + k + 3 + ... + 2k + 2k + 1 + 2k + 2 k + 1 adding and subtracting k 1 +1 13 1 1 1 > 24 + 2k + 1 + 2(k + 1) k + 1·...
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 Summer '10
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 Graph Theory, Mathematical Induction, 1 k, Natural number, 1 2k, 1 1 k, 1 1 1 k

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