Lucas Number 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:

$$ $$ $$ $$

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:

$$ $$ $$ $$

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

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