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Number Theory A proof about the fibonnaci numbers (simple for you guys)

E01

New member
Dec 8, 2013
9
The problem is as stated:
Prove that \(\displaystyle F_1*F_2+F_2*F_3+...+F_{2n-1}*F_{2n}=F^2_{2n}\)

But earlier in my text I proved by induction that \(\displaystyle F_{2n}=F_1+F_2+...+F_{2n-1}\). Do I need to use this earlier proof in my current proof. I tried adding \(\displaystyle F_{2n+1}F_{2n+2}\) to the right and left hand side of the first equation and tried to find \(\displaystyle F_{2n+1}F_{2n+2}+F^2_{2n}=F^2_{2n+2}\) but that doesn't seem to be going anywhere. (Why doesn't that seem to work in this case? Because I am multiplying two sums together?)

Am I wrong in assuming that I am supposed to prove this by induction?
 
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Petek

Member
Jan 8, 2013
20
Your equation \(\displaystyle F_{2n}=F_1+F_2+...+F_{2n-1}\) must have a typo, because it's incorrect (try it for \(\displaystyle n = 2\), say). However, I don't think you need it to solve the given problem. Proving by induction is correct, but your induction step isn't set up quite right. Assume the result is true for n and try to prove it for n + 1. Thus, we are trying to prove that

\(\displaystyle F_1F_2 + F_2F_3 + \cdots + F_{2n-1}F_{2n} + F_{2n}F_{2n+1} + F_{2n+1}F_{2n+2} = F_{2n+2}^2\)

(because \(\displaystyle 2(n+1) = 2n +2\)), and we know by induction that

\(\displaystyle F_1F_2 + F_2F_3 + \cdots + F_{2n-1}F_{2n} = F_{2n}^2\)

Can you take it from there?
 

E01

New member
Dec 8, 2013
9
Yea, after you told me how to set it up it took about ten seconds :p, and here I had sat and wondered about it for like an hour. I had assumed that I could plug N+1 into \(\displaystyle F_{2n-1}F_{2n}\) and add that back to the left hand side of the equation and get what equaled \(\displaystyle F^2_{2n+2}\). So now I know that adding the last term with n+1 substituted for n to the sum doesn't necessarily result in the actual n+1 sum as a whole. (Also thanks to you I solved the next three problems I couldn't solve :D)
 
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