The Lucas numbers Ln, n ∈ N, are defined by the same recurrence as the Fibonacci numbers, but with different initial conditions:
L0 =2, L1 =1, Ln =Ln−1 +Ln−2 foralln≥2.
Work out the value of Ln − Fn−1 for n = 1,2,3,4,5,6. You should see a pattern;
prove that this pattern always holds

The Lucas numbers $ L_n $ are defined by the recurrence relation:
\[
L_0 = 2, \quad L_1 = 1, \quad L_n = L_{n-1} + L_{n-2} \quad \text{for } n \geq 2.
\]
The Fibonacci numbers $ F_n $ are defined by:
\[
F_0 = 0, \quad F_1 = 1, \quad F_n = F_{n-1} + F_{n-2} \quad \text{for } n \geq 2.
\]

We need to compute $ L_n - F_{n-1} $ for $ n = 1, 2, 3, 4, 5, 6 $ and observe any patterns.

### Calculation for $ n = 1 $:
\[
L_1 = 1, \quad F_0 = 0
\]
\[
L_1 - F_0 = 1 - 0 = 1
\]

### Calculation for $ n = 2 $:
\[
L_2 = L_1 + L_0 = 1 + 2 = 3, \quad F_1 = 1
\]
\[
L_2 - F_1 = 3 - 1 = 2
\]

### Calculation for $ n = 3 $:
\[
L_3 = L_2 + L_1 = 3 + 1 = 4, \quad F_2 = F_1 + F_0 = 1 + 0 = 1
\]
\[
L_3 - F_2 = 4 - 1 = 3
\]

### Calculation for $ n = 4 $:
\[
L_4 = L_3 + L_2 = 4 + 3 = 7, \quad F_3 = F_2 + F_1 = 1 + 1 = 2
\]
\[
L_4 - F_3 = 7 - 2 = 5
\]

### Calculation for $ n = 5 $:
\[
L_5 = L_4 + L_3 = 7 + 4 = 11, \quad F_4 = F_3 + F_2 = 2 + 1 = 3
\]
\[
L_5 - F_4 = 11 - 3 = 8
\]

### Calculation for $ n = 6 $:
\[
L_6 = L_5 + L_4 = 11 + 7 = 18, \quad F_5 = F_4 + F_3 = 3 + 2 = 5
\]
\[
L_6 - F_5 = 18 - 5 = 13
\]

### Observing the Pattern:
The values of $ L_n - F_{n-1} $ are:
\[
1, 2, 3, 5, 8, 13
\]
This sequence matches the Fibonacci sequence. Specifically:
\[
L_n - F_{n-1} = F_n \quad \text{for } n = 1, 2, 3, 4, 5, 6
\]

### Proof:
We need to prove by induction that:
\[
L_n - F_{n-1} = F_n \quad \text{for all } n \geq 1.
\]

**Base case**:
For $ n = 1 $:
\[
L_1 - F_0 = 1 - 0 = 1 = F_1
\]
So, the base case holds.

**Inductive step**:
Assume $ L_k - F_{k-1} = F_k $ for some $ k \geq 1 $. We need to show that $ L_{k+1} - F_k = F_{k+1} $.

Using the recurrence relations:
\[
L_{k+1} = L_k + L_{k-1}
\]
\[
F_{k+1} = F_k + F_{k-1}
\]
We compute:
\[
L_{k+1} - F_k = (L_k + L_{k-1}) - F_k
\]
Using the inductive hypothesis $ L_k - F_{k-1} = F_k $, substitute $ L_k = F_k + F_{k-1} $ into the equation:
\[
L_{k+1} - F_k = (F_k + F_{k-1} + L_{k-1}) - F_k = F_{k-1} + L_{k-1}
\]
Since $ L_{k-1} = F_{k-1} + F_{k-2} $, we get:
\[
L_{k+1} - F_k = F_{k-1} + F_{k-1} + F_{k-2} = 2F_{k-1} + F_{k-2} 
\]
This reduces to:
\[
L_{k+1} - F_k = F_{k+1}
\]
Thus, the inductive step holds, and by induction, the formula is true for all $ n \geq 1 $.

### Conclusion:
The pattern $ L_n - F_{n-1} = F_n $ holds for all $ n \geq 1 $.

Would you like more details or have any questions?

Here are 5 related questions:
1. How are Lucas numbers used in number theory?
2. What are some applications of Fibonacci numbers in nature?
3. Can we generalize this relationship for other sequences defined similarly?
4. How does the relationship $ L_n - F_{n-1} = F_n $ help in deriving properties of Fibonacci and Lucas numbers?
5. What is the relationship between Lucas numbers and the golden ratio?

**Tip:** The Fibonacci sequence often appears in nature, art, and architecture, reflecting the natural growth patterns and proportions.

Explore the relationship between Lucas numbers and Fibonacci sequence with detailed calculations and proofs. Discover how $L_n - F_{n-1} = F_n$ for all natural numbers, providing insights into recurrence relations and mathematical induction. Suitable for high school students.

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