Answer
See explanation
Work Step by Step
We wish to prove by mathematical induction that for all integers \( k \ge 1 \) the sequence defined by
\[
f_1 = 1 \quad \text{and} \quad f_k = f_{k-1} + 2^k \quad \text{for } k\ge2,
\]
has the explicit formula
\[
\boxed{f_k = 2^{k+1} - 3.}
\]
---
### **Base Case** \((k = 1)\)
For \(k = 1\), the formula gives
\[
f_1 = 2^{1+1} - 3 = 2^2 - 3 = 4 - 3 = 1.
\]
This agrees with the given initial condition \(f_1 = 1\).
---
### **Inductive Step**
**Inductive Hypothesis:**
Assume that for some integer \( k \ge 1 \) the formula holds; that is, assume
\[
f_k = 2^{k+1} - 3.
\]
**To Prove:**
Show that
\[
f_{k+1} = 2^{(k+1)+1} - 3 = 2^{k+2} - 3.
\]
**Proof:**
Starting with the recursive definition,
\[
f_{k+1} = f_k + 2^{k+1}.
\]
By the inductive hypothesis, substitute \( f_k = 2^{k+1} - 3 \):
\[
f_{k+1} = \bigl(2^{k+1} - 3\bigr) + 2^{k+1} = 2 \cdot 2^{k+1} - 3 = 2^{k+2} - 3.
\]
This is exactly the formula for \(f_{k+1}\).
