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Unformatted text preview: Discrete Mathematics, Spring 2004 Homework 1 Sample Solutions Problem #1. Using induction, verify that 1(1!) + 2(2!) + ... + n ( n !) = ( n + 1)! 1 for all positive integers n . Solution : let P ( n ) be the statement 1(1!) + 2(2!) + ... + n ( n !) = ( n + 1)! 1. We prove that P ( n ) is true for all n 1 by induction on n . Base case: 1(1!) = 1, (1 + 1)! 1 = 2 1 = 1. Thus P (1) is true. Inductive step: Assume that P ( n ) is true. Then 1(1!) + 2(2!) + ... + ( n + 1)[( n + 1)!] = [1(1!) + 2(2!) + ... + n ( n !)] + ( n + 1)[( n + 1)!] = ( n + 1)! 1 + ( n + 1)[( n + 1)!] by assumption = ( n + 1)![1 + ( n + 1)] 1 = ( n + 2)! 1 . Therefore P ( n + 1) is true, and by induction the claim is proved. Problem #2. By experimenting with small values of n , guess a formula for the sum 1 1 2 + 1 2 3 + + 1 n ( n + 1) . Then use induction to verify your formula. Solution : Easy calculations show that the above sum equals 1 / 2, 2 / 3, 3 / 4 for n = 1 , 2 , 3 respectively. Therefore we guess that in general the sum equals n/ ( n +1). Here is the formal proof. Let P ( n ) be the statement 1 / (1 2) + 1 / (2 3) + + 1 / [ n ( n + 1)] = n/ ( n + 1). Clearly 1 P (1) is true, since 1 / (1 2) = 1 / 2. Now assume P ( n ) is true; then 1...
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This note was uploaded on 06/12/2011 for the course MATH 103 taught by Professor Wouters during the Spring '08 term at Wisc Oshkosh.
 Spring '08
 WOUTERS
 Algebra, Integers

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