Lucas Number 2n in terms of Square of Lucas Number n
Jump to navigation
Jump to search
Theorem
Let $L_n$ denote the $n$th Lucas number.
Then:
- $L_{2 n} = {L_n}^2 + 2 \left({-1}\right)^n$
Proof
Let:
- $\phi = \dfrac {1 + \sqrt 5} 2$
- $\hat \phi = \dfrac {1 - \sqrt 5} 2$
Note that we have:
\(\ds \phi \hat \phi\) | \(=\) | \(\ds \dfrac {1 + \sqrt 5} 2 \dfrac {1 - \sqrt 5} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {1 - 5} 4\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds -1\) |
Then:
\(\ds L_{2 n}\) | \(=\) | \(\ds \phi^{2 n} + \hat \phi^{2 n}\) | Closed Form for Lucas Numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds \phi^{2 n} + 2 \phi^n \hat \phi^n + \hat \phi^{2 n} - 2 \phi^n \hat \phi^n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\phi^n + \hat \phi^n}^2 - 2 \paren {-1}^n\) | simplifying, and from above: $\phi \hat \phi = -1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds {L_n}^2 - 2 \paren {-1}^n\) | Closed Form for Lucas Numbers |
Hence the result.
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $11$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $11$