Answer
See explanation
Work Step by Step
**Proof (by direct expansion using the Fibonacci recurrence):**
Recall the Fibonacci sequence \(\{F_n\}\) is defined by
\[
F_0 = 0,\quad F_1=1,\quad \text{and for }n\ge2:\; F_n = F_{n-1} + F_{n-2}.
\]
We want to show that for all integers \(k \ge 4\),
\[
F_k \;=\; 3\,F_{k-3} \;+\; 2\,F_{k-4}.
\]
1. **Start with the basic recurrence** \(F_k = F_{k-1} + F_{k-2}\).
2. **Express \(F_{k-1}\) in terms of earlier terms:**
\[
F_{k-1} = F_{k-2} + F_{k-3}.
\]
Substitute back into \(F_k\):
\[
F_k = \bigl(F_{k-2} + F_{k-3}\bigr) + F_{k-2}
= 2\,F_{k-2} + F_{k-3}.
\]
3. **Express \(F_{k-2}\) in terms of earlier terms:**
\[
F_{k-2} = F_{k-3} + F_{k-4}.
\]
Substitute this into the expression for \(F_k\):
\[
F_k = 2\bigl(F_{k-3} + F_{k-4}\bigr) + F_{k-3}
= 2\,F_{k-3} + 2\,F_{k-4} + F_{k-3}
= 3\,F_{k-3} \;+\; 2\,F_{k-4}.
\]
Hence, for all \(k \ge 4\),
\[
\boxed{F_k = 3\,F_{k-3} \;+\; 2\,F_{k-4}.}
\]
