# Simple Fibonacci / Lucas Numbers Relationship

Prove the identity by induction: $$F_{2n} = F_n L_n,$$ where $F_n$ and $L_n$ are the $n^{th}$ Fibonacci and Lucas number, respectively.

I have an answer but am not happy with it since it doesn't use induction. I resorted to using the identity $F_{2n} = F_{n + 1}^2 - F_{n - 1}^2$, which, admittedly, was not given to me in the text. I am, however, allowed to use the relations: $L_n = L_{n - 1} + L_{n - 2}$, $L_n = F_{n + 1} + F_{n - 1}$, and $F_{n + 1}^2 - F_n F_{n + 2} = (-1)^n$, since I proved them earlier.

So far I have shown the base cases of $n = 1, 2$, and have that $$F_{n + 1} L_{n + 1} = F_n L_n + F_{n - 1} L_{n - 1} + F_{n - 1} L_n + F_n L_{n - 1} = F_{2n} + F_{2n - 2} + F_{n - 1} L_n + F_n L_{n - 1},$$ so if I could show $F_{n - 1} L_n + F_n L_{n - 1} = 2 F_{2n - 1}$, then I would be done, but no such luck. I am trying to only use induction, no combinatorial arguments or other identities (such as Binet). Thank you in advance!

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I know you said that you don't want the Binet formula, but it yields an extremely quick way to see why the result is true - $\frac {\phi^{2n} - \phi^{-2n} } {2\sqrt{5}} = \frac {\phi^{n} - \phi^{-n}}{2 \sqrt{5} } \times ( \phi^{n} + \phi^{-n})$. It is also provides the motivation for finding related formulas, like the other one that N.S. gave in his solution. –  Calvin Lin Dec 30 '12 at 0:57
$$F_{2n} = F_n L_n \,,\mbox{and}, \, F_{n - 1} L_n + F_n L_{n - 1} = 2 F_{2n - 1}$$