Prove F_2n = FnLn

• Feb 4th 2007, 02:26 PM
yc6489
Prove F_2n = FnLn
Prove F_2n = FnLn, where Fn are fibonacci numbers and Ln are Lucas numbers.
• Feb 4th 2007, 04:01 PM
ThePerfectHacker
Quote:

Originally Posted by yc6489
Prove F_2n = FnLn, where Fn are fibonacci numbers and Ln are Lucas numbers.

That is not true consider $\displaystyle n=2$.
Then, $\displaystyle F_{2n}=3$.
$\displaystyle F_n=1$ and $\displaystyle L_n=1$.
And, $\displaystyle 1\cdot 1 \not = 2$.
• Feb 4th 2007, 08:20 PM
Soroban
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 AM
yc6489
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 PM
ThePerfectHacker
Quote:

Originally Posted by yc6489
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.)

In that case do it with induction.