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Unformatted text preview: 21-127, Concepts of Mathematics, Review Selected Solutions1. Prove the following using either weak or strong induction:(a) LetP(n) = 2nn+ 1Base Case:P(0) = 2= 10 + 1Inductive Hypothesis: AssumeP(k) is true for somekNInductive Step: We begin by citing the inductive hypothesis to conclude 2kk+ 1. We nextmultiply both sides by two to obtain: 2k+12k+ 2 =k+k+ 2. We next note that by definitionk0, and therefore 2k+1k+k+ 2k+ 2 as required.(b) Go to office hours(c) LetP(n) =nk=1k3= (nk=1k)2Base Case:P(1) = 13= 1 = 12Inductive Hypothesis: AssumeP(r) is true for somerNInductive Step: We begin by considering the quantity (r+1k=1k)2= (rk=1k+ (r+ 1))2. If welet=rk=1kthen we have (r+1k=1k)2= (+ (r+ 1))2=2+ 2(r+ 1) + (r+ 1)2. By theinductive hypothesis we know that2=rk=1k3. Looking at the other two terms, we noticethat=rk=1k=(r+1)(r)2and therefore 2(r+ 1) =r(r+ 1)2. We may finally conclude that(r+1k=1k)2= (r+ 1)2+r(r+ 1)2+rk=1k3= (r+ 1)3+rk=1k3=r+1k=1k3as 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. Whilemost problems you have done in the past always involve inducting on some obviousn, this oneis slightly more tricky in that you need to induct upon something that isnt spelt out for you.is slightly more tricky in that you need to induct upon something that isnt spelt out for you....
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This note was uploaded on 09/08/2011 for the course MATH 21-127 taught by Professor Gheorghiciuc during the Spring '07 term at Carnegie Mellon.
- Spring '07