How Can the Fibonacci Sequence Be Proved by Induction?

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The discussion revolves around proving the identity involving Fibonacci numbers: F_{1}F_{2}+F_{2}F_{3}+...+F_{2n}F_{2n+1}=F^{2}_{2n+1}-1 using mathematical induction. A participant initially attempts to prove a related identity, F_{2}+F_{4}+...+F_{2n}=F_{2n+1}-1, as a stepping stone. The conversation highlights the importance of establishing base cases and using strong induction to validate the proof. It is suggested that breaking the sum into recognizable parts can help in applying the induction hypothesis effectively. The overall consensus is that with the right approach, the proof can indeed be completed successfully.
williampb
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I've been having a lot of trouble with this proof lately:

Prove that,

F_{1}F_{2}+F_{2}F_{3}+...+F_{2n}F_{2n+1}=F^{2}_{2n+1}-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_{2}+F_{4}+...+F_{2n}=F_{2n+1}-1

And then used that in the other proof.

For F_{2}+F_{4}+...+F_{2n}=F_{2n+1}-1

First,
For n=1 (base case)

F_{2(1)}=F_{2}=1

Then,

F_{2(k+1)}=F_{2(k+1)+1}-1+F_{2(k+1)}
=(F_{2k+3}+F_{2k+2})-1
=F_{2k+5}-1 = F_{2(k+2)+1}-1

Proving that F_{2k}=F_{2k+1}-1 for all k+1.

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

F_{2n}F_{2n+1}=F^{2}_{2n+1}-1
F_{2(k+1)}F_{2(k+1)+1}=F^{2}_{2(k+1)+1}-1

Since,

F_{2(k+1)}=F_{2(k+1)+1}-1

Then the equation becomes:

F_{2(k+1)+1}F_{2(k+1)+1}-1=F^{2}_{2(k+1)+1}-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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williampb said:
I'm not sure how to prove this by straight induction
It's quite a cute proof by induction, actually, if you can prove (or already know) that \sum_{k=1}^n F_k^2 = F_n F_{n+1}.

Let's assume F_1 = F_2 = 1 and F_{n+2} = F_{n+1} + F_n for natural numbers n. Since the recursive relation refers to two lesser naturals, we should proceed with strong induction and two base cases.

Hence, for n = 1 we have:
\sum_{k=1}^{2} F_k F_{k+1} = F_1 F_2 + F_2 F_3 = 1 + 2 = 3 = 4 - 1 = 2^2 - 1 = F_3^2 - 1 = F_{2(1)+1}^2 - 1.
For n = 2 we have:
\sum_{k=1}^{4} F_k F_{k+1} = 3 + F_3 F_4 + F_4 F_5 = 3 + 6 + 15 = 24 = 25 - 1 = 5^2 - 1 = F_5^2 - 1 = F_{2(2)+1}^2 - 1.
Now suppose that \sum_{k=1}^{2j} F_k F_{k+1} = F_{2j+1}^2 - 1 for all j < n. We want to show it's true for n itself. Break up the sum into recognizable parts:
\sum_{k=1}^{2n} F_k F_{k+1} = F_1 F_2 + \sum_{k=2}^{2n} F_k F_{k+1} = 1 + \sum_{k=2}^{2n} F_k (F_k + F_{k-1}) = 1 + \sum_{k=2}^{2n} F_k^2 + \sum_{k=2}^{2n} F_k F_{k-1}.
The middle sum is 1 less than the identity at the start of my post. Note that F_k and F_{k-1} commute (luckily!) and so that last sum looks familiar, almost. Write it out in full, cut out a slice to which the induction hypothesis applies, and then deal with the rest.
 
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