Prove F_2n = FnLn, where Fn are fibonacci numbers and Ln are Lucas numbers.

Printable View

- Feb 4th 2007, 02:26 PMyc6489Prove F_2n = FnLn
Prove F_2n = FnLn, where Fn are fibonacci numbers and Ln are Lucas numbers.

- Feb 4th 2007, 04:01 PMThePerfectHacker
- Feb 4th 2007, 08:20 PMSoroban
Hello, yc6489!

Quote:

Prove: .$\displaystyle F_{2n} \:= \:F_n\!\cdot\!L_n$,

where $\displaystyle F_n$ are Fibonacci numbers and $\displaystyle L_n$ are Lucas numbers.

Are we allowed to use the*closed forms*for $\displaystyle F_n$ and $\displaystyle L_n$ ?

. . $\displaystyle F_n \;=\;\frac{(1 + \sqrt{5})^n - (1 - \sqrt{5})^n}{2^n\sqrt{5}} \qquad L_n \;=\;\frac{(1 + \sqrt{5})^n + (1 - \sqrt{5})^n}{2^n} $

Then: .$\displaystyle F_n\!\cdot\!L_n\;=\;\frac{(1 + \sqrt{5})^n - (1 - \sqrt{5})^n}{2^n\sqrt{5}}\cdot\frac{(1 + \sqrt{5})^n + (1 - \sqrt{5})^n}{2^n} $

. . $\displaystyle = \;\frac{(1 + \sqrt{5})^{2n} - (1 - \sqrt{5})^{2n}}{2^{2n}\sqrt{5}} \;=\;F_{2n} $

- Feb 5th 2007, 11:25 AMyc6489
Thanks Soroban for your reply. I think the intention of the exercise is to prove the statement from the definition of Ln = F_n-1 + F_n+1 for n>=2. (The text does not mention your particular definitions of Fn and Ln.)

- Feb 5th 2007, 12:08 PMThePerfectHacker