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


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