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Unformatted text preview: PRACTICE PROBLEMS FOR MIDTERM: SOLUTIONS The test will be given on Monday, February 1 . It will be based on Homeworks 13 (Sections 15 and 78). In preparing for the test, you can practice solving the problems from the list below. In addition, take a look at the homework problems ( at least one problem on the midterm will come from the homework ), at the examples given in the textbook, and the the examples discussed in class. 1. Prove that, for any positive integer n , ∑ n k =1 k 2 k = ( n − 1)2 n +1 + 2. The proof proceeds by induction. The base of induction is easy to verify: plugging n = 1 into our identity, we see that 1 · 2 1 = 2 = (1 − 1)2 1+1 +2. To handle the step of induction, we need to show that, for m ∈ N , ∑ m k =1 k 2 k = ( m − 1)2 m +1 +2 implies ∑ m +1 k =1 k 2 k = m 2 m +2 +2. Note that ∑ m +1 k =1 k 2 k = ∑ m k =1 k 2 k +( m +1)2 m +1 . By the induction hypothesis, ∑ m k =1 k 2 k = ( m − 1)2 m +1 + 2, hence m +1 summationdisplay k =1 k 2 k = ( m − 1)2 m +1 + 2 + ( m + 1)2 m +1 = ( ( m − 1) + ( m + 1) ) 2 m +1 + 2 = 2 m · 2 m +1 + 2 = m 2 m +2 + 2 , which is what we need. 2. The sequence ( x n ) is defined by the following rule: x 1 = 3, and x n +1 = 2 x n − 1 for n ∈ N . Prove that, for any n ∈ N , x n = 2 n + 1. Let P n be the statement that x n = 2 n + 1. We use induction to show that P n is true for any n . The basis for induction consists of verifying that P 1 is true: x 1 = 3 = 2 1 + 1. To handle the induction step, we have to show that, for any m ∈ N , P m implies P m +1 (that is, P m +1 is true whenever P m is true). By our definition, x m +1 = 2 x m − 1. By the induction hypothesis, x m = 2 m + 1, hence x m +1 = 2 x m − 1 = 2(2 m + 1) − 1 = 2 · 2 m + 2 − 1 = 2 m +1 + 1 , which is precisely what we need. 3. Suppose x and y are positive elements of an ordered field, and x 2 = y 2 . Does it follow that x = y ?...
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This note was uploaded on 05/08/2010 for the course MATH 140a taught by Professor Staff during the Winter '08 term at UC Irvine.
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