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Homework 6 Solutions W11

# Homework 6 Solutions W11 - EECS 203 Homework 6 Solutions...

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EECS 203: Homework 6 Solutions Section 4.1 1. (E) 4 a) 1 3 = (1(1 + 1) / 2) 2 b) Evaluating both sides 1=1. So True. c) The inductive hypothesis is the statement P ( k ) : 1 3 + 2 3 + ... + k 3 = k ( k +1) 2 2 . d) We need to prove that for all k 1 , given P ( k ) , the following P ( k + 1) holds: P ( k + 1) : 1 3 + 2 3 + ... + k 3 + ( k + 1) 3 = ( k +1)( k +2) 2 2 . e) Using the inductive hypothesis, P ( k ) , the left side of P ( k + 1) becomes k ( k + 1) 2 2 + ( k + 1) 3 = ( k + 1) 2 (( k 2 / 4) + k + 1) = ( k + 1) 2 ( k + 2) 2 4 = ( k + 1)( k + 2) 2 2 = P ( k + 1) which is equal to the right-handed side of P ( k + 1) . Therefore, P ( k ) P ( k + 1) . f) Since we completed the basic step and the inductive step, by principle of mathematical induction the statement is true for all positive integer n . 2. (M) 10 a) The first three sums are 1 / 2 , 2 / 3 , 3 / 4 . Looking at the pattern the sum appears to be n/ ( n + 1) . b) Let the statement P ( n ) be 1 1 · 2 + 1 2 · 3 + 1 3 · 4 + ... + 1 n · ( n +1) = n n +1 We will prove it by mathematical induction. Base step is true as 1 2 = 1 2 . Now assuming the inductive hypothesis, the inductive step is P ( k ) : 1 1 · 2 + 1 2 · 3 + 1 3 · 4 + ... + 1 k · ( k +1) + 1 ( k +1) · ( k +2) = k k +1 + 1 ( k +1) · ( k +2) = k ( k +2)+1 ( k +1)( k +2) = k 2 +2 k +1 ( k +1)( k +2) = k +1 k +2 which is same as our guess of P ( k + 1) , meaning that P ( k ) P ( k + 1) . Therefore, by principle of mathematical induction, our guess of the sum is correct. 3. (E) 18 a) P (2) is the statement that 2! < 2 2 b) The inequality in part (a) can be simplified into 2 < 4 , which means that P (2) is true. c) The inductive hypothesis is the statement P ( k ) : k ! < k k . d) In inductive step, we need to prove that if P ( k ) is true, then P ( k + 1) : ( k + 1)! < ( k + 1) k +1 is also true. e) From part (d), given that P ( k ) is true, the left side of P ( k + 1) can be derived as followed: ( k + 1)! = ( k + 1) k ! < ( k + 1) k k < ( k + 1)( k + 1) k = ( k + 1) k +1

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which concludes that P ( k + 1) : ( k + 1)! < ( k + 1) k +1 is true. Bacause of the proof of the base case and the inductive hypothesis, the statement P ( n ) is true for all integer n > 1 . 4. (E) 32 For n 1 , let P ( n ) be the statement that 3 | ( n 3 + 2 n ) or 3 divides n 3 + 2 n . BASIS STEP : P (1) is true because 3 divides 1 3 + 2 · 1 = 3 . INDUCTIVE STEP : For any k 1 , we show P ( k ) P ( k +1) . In order words, suppose that the inductive hypothesis, 3 divides k 3 +2 k , is true, we want to show that 3 also divides ( k +1) 3 +2( k +1) . If we expand the expression in question, we obtain ( k + 1) 3 + 2( k + 1) = k 3 + 3 k 2 + 3 k + 1 + 2 k + 2 = ( k 3 + 2 k ) + 3( k 2 + k + 1) By the inductive hypothesis, 3 divides k 3 + 2 k , and certainly 3 divides 3( k 2 + k + 1) . Therefore, 3 also divides their sum. Thus, P ( k ) P ( k + 1) . With proofs that P (1) and P ( k ) P ( k + 1) for all integer k 1 , it can be concluded that P ( n ) is true for all n Z + 5. (M) 44 Let P ( n )
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