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| Product of Sum and Product (Posted on 2012-06-21) |
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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.
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Submitted by Bractals
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Solution:
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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
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Comments: (
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Subject |
Author |
Date |
 | Solution | John Dounis | 2012-06-24 09:49:59 |
| Solution | John Dounis | 2012-06-24 09:43:51 |
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