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 October 9th, 2014, 03:43 AM #1 Member     Joined: May 2014 From: India Posts: 87 Thanks: 5 Math Focus: Abstract maths! A discovery in the Fibonacci sequence From a very long time, I have been interested in the mysterious properties of this seemingly simple sequence. While playing with squares I stumbled upon another two properties which thrilled me: Let F be the set of Fibonacci numbers such that, $F = \{F_n = F_{n-1} + F_{n-2}; F_0 = 0; F_1 = 1; n Ïµ N\}$ then, (i) $F_{2n} = (F_{n+1})^2 - (F_{n-1})^2$ (ii) $F_{2n+1} = (F_n)^2 + (F_{n+1})^2$ Thus far, I have never come across any website which lists these along the other properties of the Fibonacci numbers. At the same time three requests come to my mind for the great gentlemen reading this thread. (i) Help me with the proof of these properties. (ii) Enlighten me with the practical use of the Fibonacci sequence (if any)? (iii) Is there a resource(a website for instance), where we can get a summary of the properties of the Fibonacci sequence?
 October 9th, 2014, 04:16 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,657 Thanks: 2635 Math Focus: Mainly analysis and algebra Your formulae fail for $n=1$. Which $n$ do they work? Thanks from topsquark and Rishabh Last edited by v8archie; October 9th, 2014 at 04:31 AM.
 October 10th, 2014, 04:41 AM #3 Member     Joined: May 2014 From: India Posts: 87 Thanks: 5 Math Focus: Abstract maths! Thanks for pointing it out, but I think it works fine for n>1.
 October 10th, 2014, 05:37 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,657 Thanks: 2635 Math Focus: Mainly analysis and algebra They don't work for $n=2$ either. I think you have you indices wrong. Perhaps you mean $$F_{2n+1} = F_{n+1}^2 - F_{n-1}^2 \\ F_{2n} = F_{n-1}^2 + F_n^2$$ These do at least work for $n \in \{1, 2\}$. Here's a trick that might lead to the result. \begin{align*}F_n &= F_{n-1}+F_{n-2} = F_1 F_{n-1} + F_0F_{n-2} \\ &= F_1(F_{n-2} + F_{n-3}) + F_0 F_{n-2} = (F_0 + F_1) F_{n-2} + F_1 F_{n-3} \\ &= F_2 F_{n-2} + F_1 F_{n-3} \\ &\vdots \end{align*} Note that this doesn't work for the Brady numbers! Thanks from topsquark Last edited by v8archie; October 10th, 2014 at 06:07 AM.
 October 10th, 2014, 11:16 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,657 Thanks: 2635 Math Focus: Mainly analysis and algebra Note that, in the previous derivation, the indices of the first term on the right sum to the index of the term on the left. Thus, \begin{align*}F_{2n} &= F_1 F_{2n-1} + F_0 F_{2n-2} \\ &= F_2 F_{2n-2} + F_1 F_{2n-3} \\ &\vdots \\ &= F_n F_n + F_{n-1} F_{n-1} \\ &= F_n^2 + F_{n-1}^2 \\[12pt] F_{2n+1} &= F_1 F_{2n} + F_0 F_{2n-1} \\ &= F_2 F_{2n-1} + F_1 F_{2n-2} \\ &\vdots \\ &= F_n F_{n+1} + F_{n-1} F_n \\ &= (F_{n+1} - F_{n-1})F_{n+1} + F_{n-1}(F_{n+1} - F_{n-1}) \\ &= F_{n+1}^2 - F_{n-1}^2 \end{align*} Thanks from topsquark Last edited by v8archie; October 10th, 2014 at 11:18 AM.
 October 14th, 2014, 05:01 AM #6 Member     Joined: May 2014 From: India Posts: 87 Thanks: 5 Math Focus: Abstract maths! But hey, there must be some misunderstanding. Both my formulae work fine for natural numbers including 1 and 2. Given, $F_0=0; F_1=1; F_2=1; F_3=2; F_4=3; F_5=5; F_6=8; F_7=13;$ and so on... (i)$F_{2n} = (F_{n+1})^2 - (F_{n-1})^2$ When $n=1$, L.H.S. : $F_{2*1} = F_2 = 1$ R.H.S. : $(F_{1+1})^2 - (F_{1-1})^2 = {F_2}^2 - {F_0}^2 = 1^2 - 0^2 = 1$ Hence, L.H.S. = R.H.S. (ii)$F_{2n+1} = (F_n)^2 + (F_{n+1})^2$ When $n=1$, L.H.S. : $F_{2*1+1} = F_3 = 2$ R.H.S. : $(F_1)^2 + (F_{1+1})^2 = (F_1)^2 + (F_2)^2 = 1^2 + 1^2 = 2$ Hence, L.H.S. = R.H.S. Similarly, other natural number values for n also satisfies the equations. So, what now?!!!
 October 14th, 2014, 09:07 AM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,657 Thanks: 2635 Math Focus: Mainly analysis and algebra My apologies, I misread, assuming the usual $F_0 = F_1 = 1$. The proofs above can be easily adjusted for the change of index.

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