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

[E01]

The problem is as stated:
Prove that $$F_1*F_2+F_2*F_3+...+F_{2n-1}*F_{2n}=F^2_{2n}$$

But earlier in my text I proved by induction that $$F_{2n}=F_1+F_2+...+F_{2n-1}$$. Do I need to use this earlier proof in my current proof. I tried adding $$F_{2n+1}F_{2n+2}$$ to the right and left hand side of the first equation and tried to find $$F_{2n+1}F_{2n+2}+F^2_{2n}=F^2_{2n+2}$$ 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?

 

[Petek]

Your equation $$F_{2n}=F_1+F_2+...+F_{2n-1}$$ must have a typo, because it's incorrect (try it for [math] n = 2[/math], 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

[math]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[/math]

(because [math]2(n+1) = 2n +2[/math]), and we know by induction that

[math]F_1F_2 + F_2F_3 + \cdots + F_{2n-1}F_{2n} = F_{2n}^2[/math]

Can you take it from there?

 

[E01]

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 $$F_{2n-1}F_{2n}$$ and add that back to the left hand side of the equation and get what equaled $$F^2_{2n+2}$$. 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)