Prove F_2n = FnLn, where Fn are fibonacci numbers and Ln are Lucas numbers.
Hello, yc6489!
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} $