In the last section, we derived the following form of our analogue to de Moivre's theorem:
F(kn−1) + φ F(kn) =(F(n−1) + φ F(n))k
Expanding the expression on the right with the binomial formula, applying the equation
F (kn − 1) =∑0 ≤ j ≤ k ( k j ) F ( j − 1 ) F(n) j F(n−1)k−j F (kn) =∑0 ≤ j ≤ k ( k j ) F ( j ) F(n) j F(n−1)k−j
Combine these two formulas with the basic recurrence relation for the Fibonacci numbers to yield:
F (kn + m) =∑0 ≤ j ≤ k ( k j ) F ( j + m ) F(n) j F(n−1)k−j
Using the fact that
L (kn + m) =∑0 ≤ j ≤ k ( k j ) L ( j + m ) F(n) j F(n−1)k−j
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