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could someone possibly help me with a proof.

prove $a_n = F_{2n-1}$

for fibonacci numbers and a recurrence relation where

$a_1 = 1$

$a_2 = 2$

$a_3 = 5$

$a_4 = 13$

$a_5 = 34$


$$ a_{n+2}a_n = a^2_{n+1} +1. $$

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closed as unclear what you're asking by Martin Sleziak, Danny Cheuk, dfeuer, Nick Peterson, Cameron Buie Sep 19 '13 at 4:33

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What is $a_n$ ? –  Amr Nov 20 '12 at 23:04
What is the recurrence relation? –  Mark Bennet Nov 20 '12 at 23:07
Note, as I have attempted to get across any number of times, the set of sequences of, say, rational numbers such that $ x_{n+2} = 3 x_{n+1} - x_n $ make up a vector space over $\mathbb Q.$ Furthermore the dimension of this vector space is two, so a basis is simply two such sequences, carefully indexed, so that neither is a constant multiple of the other. –  Will Jagy Nov 20 '12 at 23:21
possible duplicate of Recurrence relation, Fibonacci numbers –  Micah Nov 21 '12 at 0:00
I put a proof that $a_{n+2} = 3 a_{n+1} - a_n$ for the above sequence of odd-index Fibonacci numbers here: math.stackexchange.com/questions/241663/… –  Will Jagy Nov 21 '12 at 3:18

1 Answer 1

HINT: This amounts to showing that $F_{2n+3}F_{2n-1}=F_{2n+1}^2+1$, which is a slightly rearranged version of one case of Catalan’s identity.

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Cute. +1 (Nice that there's no software filter for short comments.)(...which this no longer is.) –  Rick Decker Nov 21 '12 at 1:29

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