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Sequence/Induction question

  • Thread starter nugget
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  • #1
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Homework Statement



A sequence xn is defined recursively by x1 = 2 and xn+1 = 8xn-1/4xn+3 for n ∈ N.
a. Prove by induction or otherwise that xn -1 > 0 for all n ∈ N.

Homework Equations



given...

The Attempt at a Solution



x1 is given as 2, therefore x1 - 1 = 1 > 0, which satisfies the criteria.

I'm not exactly sure how to employ induction (or the 'otherwise' method), from what I understand I need to prove that xn+1 - 1 > 0, as then the base case (n=1) and all other cases are proved... on the right track?
 

Answers and Replies

  • #2
LCKurtz
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Is that middle term -(1/4)xn or -1/(4xn)?
 
  • #3
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Yes, you seem to be on the right track.

To use induction, you need to prove two things:

1) that your statement is true for a particular case.
2) that your statement being true for one case implies that it is true for the next case.

I assume you meant to write (8xn - 1)/(4xn + 3) there.

So 1) is easy, you already proved it.
For 2), can you show that if xn > 1, xn+1 > 1 for all n [tex]\in[/tex] N?
 

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