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!
Best Answer
Try to prove induction both relations at once:
$$F_{2n} = F_n L_n \,,\mbox{and}, \, F_{n - 1} L_n + F_n L_{n - 1} = 2 F_{2n - 1}$$