- #1
williampb
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I've been having a lot of trouble with this proof lately:
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.
For F[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 that F[itex]_{2k}[/itex]=F[itex]_{2k+1}[/itex]-1 for 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.
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.
For F[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 that F[itex]_{2k}[/itex]=F[itex]_{2k+1}[/itex]-1 for 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.