Fibonacci Number of Index 3n as Sum of Cubes of Fibonacci Numbers
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Theorem
Let $F_n$ denote the $n$th Fibonacci number.
Then:
- $F_{3 n} = {F_{n + 1} }^3 + {F_n}^3 - {F_{n - 1} }^3$
Proof
From Honsberger's Identity:
- $\forall m, n \in \Z_{>0}: F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$
Setting $m = 2 n$:
\(\ds F_{3 n}\) | \(=\) | \(\ds F_{2 n + n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds F_{2 n - 1} F_n + F_{2 n} F_{n + 1}\) | Honsberger's Identity | |||||||||||
\(\ds \) | \(=\) | \(\ds F_{n + \paren{n - 1} } F_n + F_{2 n} F_{n + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren { {F_{n - 1} }^2 + {F_n}^2} F_n + F_{2 n} F_{n + 1}\) | Honsberger's Identity | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren { {F_{n - 1} }^2 + {F_n}^2} F_n + \paren { {F_{n + 1} }^2 - {F_{n - 1} }^2} F_{n + 1}\) | Fibonacci Number of Index 2n as Sum of Squares of Fibonacci Numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds {F_{n - 1} }^2 F_n + {F_n}^3 + {F_{n + 1} }^3 - {F_{n - 1} }^2 F_{n + 1}\) | rearranging | |||||||||||
\(\ds \) | \(=\) | \(\ds {F_{n - 1} }^2 F_n + {F_n}^3 + {F_{n + 1} }^3 - {F_{n - 1} }^2 \paren {F_n + F_{n - 1} }\) | Definition of Fibonacci Number | |||||||||||
\(\ds \) | \(=\) | \(\ds {F_{n + 1} }^3 + {F_n}^3 - {F_{n - 1} }^3\) | simplifying |
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $5$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $5$