Proving the Recursive Fibonacci Problem: Is (Fn+1+Fn-1)Fn always a Fibonacci?

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Homework Statement



Is: (Fn+1+Fn-1)Fn always a Fibonacci?

The Attempt at a Solution


I have no clue!
I know I'm supposed to show work and all but I'm so lost, any direction would be appreciated. Even if you just give me a link to read.
 
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hi swtlilsoni! :smile:
swtlilsoni said:

Homework Statement



Is: (Fn+1+Fn-1)Fn always a Fibonacci?

The Attempt at a Solution


I have no clue!
I know I'm supposed to show work and all but I'm so lost, any direction would be appreciated. Even if you just give me a link to read.

can you start by defining what a Fibonacci is?

(and please check the question … it doesn't look right :redface:)
 
hi swtlilsoni! :wink:

(just got up :zzz: …)

i suggest you look at http://en.wikipedia.org/wiki/Fibonacci_Number" instead, and work through the various proofs :smile:
 
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good morning!
okay so it seems like these proofs are to prove a given number is fibonacci?
so do you mean I should solve for the number in terms of n, then try to use one of those proofs to show it is fibonacci?
 
hi swtlilsoni! :smile:

if you look down the page, under "Fifth identity", you can see that it equals F2n

(it's also equation 27 in http://mathworld.wolfram.com/FibonacciNumber.html" )
swtlilsoni said:
… so do you mean I should solve for the number in terms of n, then try to use one of those proofs to show it is fibonacci?

no, i mean you should try to do it yourself, using the techniques in the other proofs as examples to help you :smile:
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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