Answer
See explanation
Work Step by Step
## Statement
We wish to prove:
\[
F_n < 2^n \quad \text{for all integers }n \ge 1.
\]
---
## Proof by Strong Induction
### 1. Base Cases
**\(n=1\)**: \(F_1 = 1 < 2^1 = 2.\)
**\(n=2\)**: \(F_2 = 1 < 2^2 = 4.\)
Thus, the statement holds for \(n=1\) and \(n=2\).
### 2. Inductive Hypothesis
Assume for some integer \(n \ge 2\) that for **all** integers \(k\) with \(1 \le k \le n\), we have
\[
F_k < 2^k.
\]
We must prove \(F_{n+1} < 2^{n+1}\).
### 3. Inductive Step
Using the Fibonacci recurrence:
\[
F_{n+1} = F_n + F_{n-1}.
\]
By the strong inductive hypothesis:
\(F_n < 2^n\).
\(F_{n-1} < 2^{n-1}\).
Hence,
\[
F_{n+1}
\;=\;
F_n + F_{n-1}
\;<\;
2^n + 2^{n-1}
\;=\;
2^{n-1}(2 + 1)
\;=\;
3 \cdot 2^{n-1}.
\]
Finally, note that \(3 \cdot 2^{n-1} < 4 \cdot 2^{n-1} = 2^{n+1}\) because \(3 < 4\). Therefore,
\[
F_{n+1} < 2^{n+1}.
\]
This completes the strong inductive step.
---
## Conclusion
By strong induction, for all integers \(n \ge 1\),
\[
\boxed{F_n < 2^n}.
\]
