Midterm1Solutions

Midterm1Solutions - Faculty of Mathematics University of...

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Faculty of Mathematics University of Waterloo MATH 135 MIDTERM EXAM #1 Fall 2007 Monday 15 October 2007 19:00 – 20:15 Solutions 1. Let a, b, c Z . Suppose that P is the statement: “If a | b or a | c , then a | bc ”. (a) Write down the converse of P . [2] Solution : The converse is “If a | bc , then a | b or a | c .” (b) Disprove the converse of P . [2] Solution : Suppose a = 4, b = 2 and c = 2. Then a | bc , since 4 | 4, but a 6 | b and a 6 | c since 4 6 | 2. Thus, a = 4, b = 2 and c = 2 is a counterexample that disproves the converse of P . (c) Write down the contrapositive of P . [2] Solution : The contrapositive is “If a 6 | bc , then a 6 | b and a 6 | c .” 2. Suppose that x 1 = 10, x 2 = 16 and x n = 4 x n - 1 - 4 x n - 2 for n 3. [8] Prove that x n = 3(2 n +1 ) - n 2 n for all n P . Solution : We prove the result by strong induction on n . Base Cases n = 1: x 1 = 10 and 3(2 2 ) - 1 · 2 1 = 3(4) - 2 = 10 n = 2: x 2 = 16 and 3(2 3 ) - 2 · 2 2 = 3(8) - 2(4) = 16 Therefore, the formula holds for n = 1 and n = 2. Induction Hypothesis Assume that the result is true for n = 1 , 2 , . . . , k for some k P , k 2. That is assume, that x n = 3(2 n +1 ) - n 2 n for n = 1 , 2 , . . . , k .
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MATH 135, Midterm #1 Solutions Page 2 of 5 Induction Conclusion Consider n = k + 1. Then x k +1 = 4 x k - 4 x k - 1 = 4(3(2 k +1 ) - k 2 k ) - 4(3(2 k ) - ( k - 1)2 k - 1 ) (by Induction Hypothesis) = 12(2 k +1 ) - 12(2 k ) - 4 k 2 k + 4( k - 1)2 k - 1 = 2 k (24 - 12) - 2 k - 1 (4 k (2) - 4( k - 1)) = 2 k (12) - 2 k - 1 (4 k + 4) = 2 k (2 2 )(3) - 2 k - 1 (2 2 )( k + 1) = 3(2 k +2 ) - ( k
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Midterm1Solutions - Faculty of Mathematics University of...

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