Prove F_2n = FnLn

**yc6489** · February 4th 2007, 02:26 PM

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

**ThePerfectHacker** · February 4th 2007, 04:01 PM

Originally Posted by yc6489

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

That is not true consider $n=2$ .
Then, $F_{2n}=3$ .
$F_n=1$ and $L_n=1$ .
And, $1\cdot 1 \not = 2$ .

**Soroban** · February 4th 2007, 08:20 PM

Hello, yc6489!

Prove: . $F_{2n} \:= \:F_n\!\cdot\!L_n$ ,
where $F_n$ are Fibonacci numbers and $L_n$ are Lucas numbers.

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

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

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

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

**yc6489** · February 5th 2007, 11:25 AM

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.)

**ThePerfectHacker** · February 5th 2007, 12:08 PM

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.

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