Using induction, prove that any non-negative integer can be written as the sum of distinct non-consecutive Lucas numbers.

Steps to Solve

1. Base Case: Verifying for n = 0 and n = 1

   The Lucas numbers are defined as follows:

   • $L_0 = 2$, $L_1 = 1$, $L_2 = 3$, $L_3 = 4$, $L_4 = 7$, $L_5 = 11$, $L_6 = 18$, ...

   For $n = 0$: We can express $0$ as the sum of no Lucas numbers, satisfying the condition.
   For $n = 1$: We can express $1$ as $L_1$.
   Thus, the base cases hold.

2. Inductive Hypothesis: Assume the statement is true for all integers up to k

   Assume that any non-negative integer $k$ can be expressed as the sum of distinct non-consecutive Lucas numbers. We need to prove it holds for $k + 1$.

3. Inductive Step: Prove for k + 1

   Consider $k + 1$. There are two cases:

   • If $k + 1$ is equal to a Lucas number, say $L_m$, then it can be represented as itself.
   • If $k + 1$ is not a Lucas number, then consider the largest Lucas number less than or equal to $k + 1$, denoted as $L_m$.

   We can express $k + 1$ as: $$ k + 1 = (k + 1 - L_m) + L_m $$

   By the inductive hypothesis, $(k + 1 - L_m)$ can be expressed as the sum of distinct non-consecutive Lucas numbers. Since $L_{m-1}$ and $L_{m-2}$ are non-consecutive to $L_m$, their inclusion is valid.

4. Conclusion: Combining parts to prove

   Thus, we have shown that $k + 1$ can also be represented as the sum of distinct non-consecutive Lucas numbers. Therefore, by the principle of mathematical induction, the statement holds for all non-negative integers.