Answer
See explanation
Work Step by Step
We wish to prove by mathematical induction that for all integers \( k \ge 0 \) the sequence defined by
\[
h_0 = 1 \quad\text{and}\quad h_k = 2^k - h_{k-1} \quad \text{for } k \ge 1,
\]
has the closed-form solution
\[
\boxed{h_k = \frac{2^{k+1} + (-1)^k}{3}}.
\]
---
### **Base Case** \((k = 0)\)
For \( k = 0 \), the formula gives
\[
h_0 = \frac{2^{0+1} + (-1)^0}{3} = \frac{2 + 1}{3} = \frac{3}{3} = 1.
\]
This agrees with the given initial condition \( h_0 = 1 \).
---
### **Inductive Step**
**Inductive Hypothesis:**
Assume that for some integer \( k \ge 1 \), the formula holds for \( k-1 \); that is, assume
\[
h_{k-1} = \frac{2^{k} + (-1)^{k-1}}{3}.
\]
**Goal:**
Show that under this assumption, the formula also holds for \( h_k \); namely, that
\[
h_k = \frac{2^{k+1} + (-1)^k}{3}.
\]
**Proof:**
Starting with the recursive definition, we have
\[
h_k = 2^k - h_{k-1}.
\]
Substitute the inductive hypothesis for \( h_{k-1} \):
\[
h_k = 2^k - \frac{2^k + (-1)^{k-1}}{3}.
\]
Write \(2^k\) as \(\frac{3 \cdot 2^k}{3}\) to combine terms:
\[
h_k = \frac{3 \cdot 2^k - \bigl(2^k + (-1)^{k-1}\bigr)}{3} = \frac{3 \cdot 2^k - 2^k - (-1)^{k-1}}{3}.
\]
Simplify the numerator:
\[
3 \cdot 2^k - 2^k = 2 \cdot 2^k = 2^{k+1},
\]
so
\[
h_k = \frac{2^{k+1} - (-1)^{k-1}}{3}.
\]
Notice that
\[
-(-1)^{k-1} = (-1)^k,
\]
so the expression becomes
\[
h_k = \frac{2^{k+1} + (-1)^k}{3}.
\]
This is exactly the desired formula for \( h_k \).
