Lucas Numbers as Sum of Fibonacci Numbers

Theorem

Let $L_k$ be the $k$th Lucas number.

Then:

$L_n = F_{n-1} + F_{n+1}$

where $F_k$ is the $k$th Fibonacci number.

Proof

Proof by induction:

For all $n \in \N, n \ge 1$, let $P \left({n}\right)$ be the proposition $L_n = F_{n-1} + F_{n+1}$.

Basis for the Induction

• $P(1)$ is true, as this just says $L_1 = 1 = F_0 + F_2$ from the definition of the Fibonacci numbers.

This is our basis for the induction.

Induction Hypothesis

• Let us make the supposition that, for some $k \in \N: k \ge 1$, the proposition $P \left({j}\right)$ holds for all $j \in \N: 1 \le j \le k$.

We shall show that it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

$L_k = F_{k-1} + F_{k+1}$

Then we need to show:

$L_{k+1} = F_k + F_{k+2}$

Induction Step

This is our induction step:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle L_{k+1}\)	\(=\)	\(\displaystyle L_{k-1} + L_k\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle F_{k-2} + F_k + F_{k-1} + F_{k+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by the induction hypothesis
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({F_{k-2} + F_{k-1} }\right) + \left({F_k + F_{k+1} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle F_k + F_{k+2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

So $P \left({k+1}\right)$ and the result follows by the Second Principle of Mathematical Induction.

Therefore:

$\forall n \in \N, n \ge 1: L_n = F_{n-1} + F_{n+1}$

Note

Some treatments use this relationship as the definition of the Lucas numbers, and from it deduce the recurrence relation $L_n = L_{n-2} + L_{n-1}$.

The proof follows by induction, as follows.

We have that:

• $L_1 = F_0 + F_2 = 0 + 1 = 1$
• $L_2 = F_1 + F_3 = 1 + 2 = 3$
• $L_3 = F_2 + F_4 = 1 + 3 = 4$.

So for $n = 3$, the recurrence relation holds, as $L_3 = L_1 + L_2$.

Now suppose that $L_k = L_{k-2} + L_{k-1}$.

We have that:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle L_{k+1}\)	\(=\)	\(\displaystyle F_k + F_{k+2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle F_{k-2} + F_{k-1} + F_k + F_{k+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({F_{k-2} + F_k}\right) + \left({F_{k-1} + F_{k+1} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle L_{k-1} + L_k\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Hence $L_n = L_{n-2} + L_{n-1}$ follows by the Second Principle of Mathematical Induction, as before.

$\blacksquare$

The difference here is that $L_0$ is not specifically defined, but is left to follow of its own accord from $L_0 + L_1 = L_2$.

Sources

• George E. Andrews: Number Theory (1971): $\S 1.1$: Exercise $13$