I've been having a lot of trouble with this proof lately:(adsbygoogle = window.adsbygoogle || []).push({});

Prove that,

F[itex]_{1}[/itex]F[itex]_{2}[/itex]+F[itex]_{2}[/itex]F[itex]_{3}[/itex]+...+F[itex]_{2n}[/itex]F[itex]_{2n+1}[/itex]=F[itex]^{2}_{2n+1}[/itex]-1

Where the subscript denotes which Fibonacci number it is. I'm not sure how to prove this by straight induction so what I did was first prove that,

F[itex]_{2}[/itex]+F[itex]_{4}[/itex]+...+F[itex]_{2n}[/itex]=F[itex]_{2n+1}[/itex]-1

And then used that in the other proof.

ForF[itex]_{2}[/itex]+F[itex]_{4}[/itex]+...+F[itex]_{2n}[/itex]=F[itex]_{2n+1}[/itex]-1

First,

For n=1 (base case)

F[itex]_{2(1)}[/itex]=F[itex]_{2}[/itex]=1

Then,

F[itex]_{2(k+1)}[/itex]=F[itex]_{2(k+1)+1}[/itex]-1+F[itex]_{2(k+1)}[/itex]

=(F[itex]_{2k+3}[/itex]+F[itex]_{2k+2}[/itex])-1

=F[itex]_{2k+5}[/itex]-1 = F[itex]_{2(k+2)+1}[/itex]-1

Proving thatF[itex]_{2k}[/itex]=F[itex]_{2k+1}[/itex]-1for all k+1.

Now my question is, would this be a valid method for the proof first stated:

F[itex]_{2n}[/itex]F[itex]_{2n+1}[/itex]=F[itex]^{2}_{2n+1}[/itex]-1

F[itex]_{2(k+1)}[/itex]F[itex]_{2(k+1)+1}[/itex]=F[itex]^{2}_{2(k+1)+1}[/itex]-1

Since,

F[itex]_{2(k+1)}[/itex]=F[itex]_{2(k+1)+1}[/itex]-1

Then the equation becomes:

F[itex]_{2(k+1)+1}[/itex]F[itex]_{2(k+1)+1}[/itex]-1=F[itex]^{2}_{2(k+1)+1}[/itex]-1

Something doesn't sit right with me. I feel like this is incorrect. If it is, any and all help would be appreciated. Thank you.

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# Fibonacci Proof by Induction

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