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Product of Sum and Product (Posted on 2012-06-21) Difficulty: 3 of 5


Let {an} be a sequence of real numbers defined by

    a0 ∉ {0,1},
    a1 = 1 - a0, and
    an+1 = 1 - an(1 - an) for all n ≥ 1.

Let Pn and Sn be defined by

    Pn = a0a1a2 ··· an and

    Sn = 1/a0 + 1/a1 + 1/a2 + ··· + 1/an

for all n ≥ 0.

Prove the following

    PnSn = 1 for all n ≥ 0.

  Submitted by Bractals    
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Solution: (Hide)


Lemma:  an+1 + Pn = 1  [∀n≥0]

              Basis:  a1 + Po = a1 + a0 = 1

          Induction:  an+1 + Pn = 1  ⇒  an+2 + Pn+1 = 1 [∀n≥0]

                          an+2 + Pn+1 = [1 - an+1(1 - an+1)] + Pnan+1
                                     = 1 - an+1[1 - an+1 - Pn]
                                     = 1 - an+1[1 - (an+1 + Pn)]
                                     = 1 - an+1[1 - 1]
                                     = 1

QED

Problem:  PnSn = 1  [∀n≥0]


              Basis:  P0S0 = (a0)(1/a0) = 1

          Induction:  PnSn = 1  ⇒  Pn+1Sn+1 = 1  [∀n≥0]

                          Pn+1Sn+1 = (Pnan+1)(Sn + 1/an+1)
                                   = (PnSn)an+1 + (Pnan+1)(1/an+1)
                                   = an+1 + Pn
                                   = 1

QED

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionSolutionJohn Dounis2012-06-24 09:49:59
SolutionSolutionJohn Dounis2012-06-24 09:43:51
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