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Unformatted text preview: SOLUTION SET 5—DUE 3/5/2008 Please report any errors in this document to Ian Sammis ( isammis@math.berkeley.edu ). For more practice try odd problems. Problem 1 (#4.1.16) . Prove that for every positive integer n , 1 · 2 · 3 + 2 · 3 · 4 + ··· + n ( n + 1)( n + 2) = n ( n + 1)( n + 2)( n + 3) / 4 Proof. Let P ( n ) be the equation as shown above. First, the base step: 1 · 2 · 3 = 1 · 2 · 3 · 4 / 4 = 6 so P (1) is true. Now, suppose P ( n ) is true. Then 1 · 2 · 3 + ··· + n ( n + 1)( n + 2) + ( n + 1)( n + 2)( n + 3) = n ( n + 1)( n + 2)( n + 3) 4 + ( n + 1)( n + 2)( n + 3) =( n + 1)( n + 2)( n + 3) h n 4 + 1 i =( n + 1)( n + 2)( n + 3) n + 4 4 = ( n + 1)( n + 2)( n + 3)( n + 4) 4 Since P (1) is true and P ( n ) ⇒ P ( n +1), we know that P ( n ) is true for all n ∈ N . Problem 2 (#4.1.18) . Let P ( n ) be the statement that n ! < n n , where n is an integer greater than 1. a) What is the statement P (2) . b) Show that P (2) is true, completing the basis step of the proof. c) What is the inductive hypothesis? d) What do you need to prove in the inductive step? e) Complete the inductive step. f) Explain why these steps show that this inequality is true whenever n is an integer greater than 1. Solution. a) P (2) says 2! < 2 2 . b) Since 2! = 2, 2 2 = 4, and 2 < 4, P (2) is true. c) The inductive hypothesis here is: Suppose P ( n ) (that is, suppose n ! < n n ). d) To complete the inductive step, one must prove that the inductive hypothesis implies P ( n + 1) (that is, that ( n + 1)! < ( n + 1) n +1 ). e) From the inductive hypothesis, we know n ! < n n . Multiplying both sides by n +1, we see ( n +1)! < ( n +1) n n . Since n < n +1, and since n > 1, n n < ( n +1) n , so ( n + 1)! < ( n + 1) n +1 . Thus, P ( n ) ⇒ P ( n + 1). f) Suppose there were some positive integer n > 2 for which P ( n ) were false. Then, since the integers are wellordered, there must be a least such n . But then P ( n 1) must have been true, so P ( n ) must have been as well (by modus ponens , 1 2 SOLUTION SET 5—DUE 3/5/2008 since P ( n 1) and P ( n 1) ⇒ P ( n )). Contradiction, so P ( n ) must be true for all positive integers greater than or equal to 2. Problem 3 (#4.1.36) . Prove that 21 divides 4 n +1 +5 2 n 1 whenever n is a positive integer. Proof. First, the base step: For n = 1, 4 2 + 5 = 21, and certainly 21  21. Now, suppose 21  ( 4 n +1 + 5 2 n 1 ) . Consider 4 n +2 + 5 2 n +1 . Factoring, we see that this is equal to 4(4 n +1 ) + 25(5 2 n 1 ). We need the form from the induction hypothesis, so let’s add it in the cheesiest way possible: 4(4 n +1 ) + 25(5 2 n 1 ) =4(4 n +1 + 5 2 n 1 5 2 n 1 ) + 25 * 5 2 n 1 =4(4 n +1 + 5 2 n 1 ) + (25 4)5 2 n 1 =4(4 n +1 + 5 2 n 1 ) + (21)5 2 n 1 21 divides the first term by the induction hypothesis, and the second term be cause it’s manifestly a multiple of 21. Thus, 21  4 n +2 + 5 2 n +1 , completing the induction step. Thus, 21 divides 4 n +1 + 5 2 n 1 whenever n is a positive integer, as...
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This note was uploaded on 12/07/2011 for the course MATH MATH192 taught by Professor Michealalain during the Spring '11 term at Université Joseph Fourier Grenoble I.
 Spring '11
 MichealAlain
 Math

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