# 1 - P m,P m 1 of propositions is true provided(i P m is...

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Math 310 - hw 1 solutions Wednesday, 2 Sept 2009 1.4 (a) Guess a formula for 1 + 3 + ··· + (2 n - 1) by evaluating the sum for n = 1 , 2 , 3 and 4. [For n = 1, the sum is simply 1.] We evaluate the sums for n = 1 , 2 , 3 and 4 as follows: 1 = 1 1 + 3 = 4 1 + 3 + 5 = 9 1 + 3 + 5 + 7 = 16 In each case the sum is n 2 so we conjecture that 1 + 3 + ··· + (2 n - 1) = n 2 for all n N . ± (b) Prove your formula using mathematical induction. For a natural number n let P n denote the statement 1 + 3 + ··· + (2 n - 1) = n 2 . Since 1 = 1 2 , the base case P 1 holds. Now suppose that P n holds for some n N . We verify that P n +1 holds 1 + 3 + ··· + (2 n - 1) + (2( n + 1) - 1) = n 2 + (2( n + 1) - 1) by P n = n 2 + 2 n + 1 = ( n + 1) 2 . Hence, P n +1 holds (by transitivity of equality). Thus for all n N , P n implies P n +1 . By the principle of mathematical induction we conclude that P n is true for all n N . ± 1.8 The principle of mathematical induction can be extended as follows. A list
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Unformatted text preview: P m ,P m +1 ,... of propositions is true provided (i) P m is true, (ii) P n +1 is true whenever P n is true and n ≥ m . (a) Prove that n 2 > n + 1 for all integers n ≥ 2. For a natural number n let P n denote the inequality n 2 > n + 1. Now the base case is m = 2, so we must verify P 2 . For n = 2, n 2 = 2 2 = 4, while n + 1 = 2 + 1 = 3: since 4 > 3, P 2 holds. Now suppose that P n holds for some integer n ≥ 2. We must check that P n +1 also holds: ( n + 1) 2 = n 2 + 2 n + 1 > ( n + 1) + 2 n + 1 by P n = 3 n + 2 > n + 2 since n ≥ 2 > . Hence, ( n + 1) 2 > ( n + 1) + 1. Thus for every integer n ≥ 2, P n implies P n +1 . By the principle of mathe-matical induction we conclude that P n is true for every integer n ≥ 2. ±...
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