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Exam2PracticeSoln

# Exam2PracticeSoln - 21-127 Concepts of Mathematics Review...

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21-127, Concepts of Mathematics, Review Selected Solutions 1. Prove the following using either weak or strong induction: (a) Let P ( n ) = 2 n n + 1 Base Case: P (0) = 2 0 = 1 0 + 1 Inductive Hypothesis: Assume P ( k ) is true for some k N Inductive Step: We begin by citing the inductive hypothesis to conclude 2 k k + 1. We next multiply both sides by two to obtain: 2 k +1 2 k + 2 = k + k + 2. We next note that by definition k 0, and therefore 2 k +1 k + k + 2 k + 2 as required. (b) Go to office hours (c) Let P ( n ) = n k =1 k 3 = ( n k =1 k ) 2 Base Case: P (1) = 1 3 = 1 = 1 2 Inductive Hypothesis: Assume P ( r ) is true for some r N Inductive Step: We begin by considering the quantity ( r +1 k =1 k ) 2 = ( r k =1 k + ( r + 1)) 2 . If we let α = r k =1 k then we have ( r +1 k =1 k ) 2 = ( α + ( r + 1)) 2 = α 2 + 2 α ( r + 1) + ( r + 1) 2 . By the inductive hypothesis we know that α 2 = r k =1 k 3 . Looking at the other two terms, we notice that α = r k =1 k = ( r +1)( r ) 2 and therefore 2 α ( r + 1) = r ( r + 1) 2 . We may finally conclude that ( r +1 k =1 k ) 2 = ( r + 1) 2 + r ( r + 1) 2 + r k =1 k 3 = ( r + 1) 3 + r k =1 k 3 = r +1 k =1 k 3 as desired. (d) Go to office hours (e) Before we begin, it is worth noting that the trick in this problem is inducting upon a+b. While

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Exam2PracticeSoln - 21-127 Concepts of Mathematics Review...

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