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In a recent post I used the Fibonacci tiling to demonstrate that

$$F_nF_{n+1}=\sum_{k=0}^nF_k^2$$

Having some additional Fibonacci tilings in my toolbox (see two figures below), I set out to see what else I could find. Thus, I arrived at

$$ F_{n~(\text{odd})}^2=\sum_{k=0}^{n-1}F_kF_{k+1}+1\\ F_{n~(\text{even})}^2=\sum_{k=0}^{n-1}F_kF_{k+1} $$

These equations can be found online and in the literature (see References). My question is how to derive them from the basic Fibonacci identities. Various attempts rapidly got me nowhere. Specifically, I tried starting with the recursion relation and Cassini's identity.

In a similar vein, using golden ratio tilings, I found the remarkably symmteric relations

$$ \varphi^n\varphi^{n+1}=\sum_{k=-\infty}^{n}(\varphi^k)^2\\ (\varphi^n)^2=\sum_{k=-\infty}^{n-1}\varphi^k\varphi^{k+1} $$

Here again, I'm at a loss as to how to derive these results. An interesting result accrues if we substitute the first of the above in the second, to wit

$$ (\varphi^n)^2=\sum_{k=-\infty}^{n-1}(n-k)(\varphi^k)^2 $$

References

S Vajda, Fibonacci and Lucas numbers, and the Golden Section: Theory and Applications, Dover Press (2008).

R Knott, Fibonacci and Golden Ratio Formulae

Figures

Fibonacci odd

Fibonacci even

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in induction on n: $\sum _0 ^0 (f_n ^ 2)$ = $f_0 ^ 2$ = 1 = $f_0$$f_1$ - checked

$f_n+1$$f_n+2$ = $f_n ^2$+$f_n$$f_n+1$ = $\sum _0^ n$$^+$$^1$$ f_n^2$

WNTP.

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  • $\begingroup$ same for golden ratio $\endgroup$
    – DIexp
    Oct 2 '17 at 8:04
  • $\begingroup$ My error in thinking was in trying to do the induction starting at $n$ and working backwards. Rue the day. BTW what is WNTP? $\endgroup$ Oct 2 '17 at 19:05
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The Fibonacci number identities can be derived by induction using the fact that $$F_{n+1}^2 - F_{n-1}^2 = (F_{n+1} - F_{n-1}) (F_{n+1} + F_{n-1}) = F_n(F_{n+1} + F_{n-1}) = F_{n-1} F_n + F_n F_{n+1}.$$ Similarly you can prove the identities for the golden ratio by induction.

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  • $\begingroup$ I appreciate your response, but you are going to have to me considerably more than that. The derivation that I see in Vajda, which I received after the posting, is a lot more involved and uses something called a generalized Fibonacci sequence. $\endgroup$ Oct 1 '17 at 20:34
  • $\begingroup$ @CyeWaldman Is this answer unclear? I don't understand. $\endgroup$
    – user486572
    Oct 2 '17 at 0:29
  • $\begingroup$ Not at all, I just meant to show me how to do it. I don't see a clear path to the solution. $\endgroup$ Oct 2 '17 at 2:31

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