Geometric sequences and Fibbonacci Numbers

  • Thread starter armolinasf
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  • #1
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Homework Statement



A) In a certain geometric sequence every term is the sum of the two preceding terms, viz. the Fibonacci sequence, what can be said about the common ratio of the sequence?

So how do I go from 1,1,2,3,5,8,13,21,34... to (1+/-sqrt(5))/2?

Then find numbers A and B such (for all n) the nth term of the Fibonacci sequence is equal to:

A((1+sqrt(5))/2)^n + B((1-sqrt(5))/2)^n



B) The first term of a geometric sequence is 1 and the fourth term is a>0. Find the second and third terms.



2. The attempt at a solution

A) Unfortunately, I'm not even sure where to begin...

B) I know that the answer is a^(1/3) and a^(2/3). But figured it out just by thinking about the geometric mean being sqrt(ab), so I'm wondering how it would be done analytically
 
Last edited:

Answers and Replies

  • #2
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B)
You're given 1, ___, ___, a, ...

So you know that 1*r*r*r =a Then, r^3=a, r=a^(1/3) and your result follows.
 
  • #3
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So its basically just applying some logic. Any ideas on the other problem, thanks.
 

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