Answer
See explanation
Work Step by Step
Below is a brief outline of how to prove, by induction, that for all integers \(k \ge 0\),
\[
h_k \;=\; \frac{2^{\,k+1} + (-1)^k}{3}
\]
satisfies the recursion
\[
h_k = 2^k - h_{k-1}
\quad\text{with}\quad
h_0 = 1.
\]
---
### **Base Case** \((k=0)\)
We check the formula for \(k=0\):
\[
h_0 \;=\; \frac{2^{0+1} + (-1)^0}{3}
\;=\; \frac{2 + 1}{3}
\;=\; 1.
\]
This matches the given initial condition \(h_0 = 1\). So the base case holds.
---
### **Inductive Step**
Assume that for some \(k \ge 1\) the formula holds at \(k-1\), i.e.,
\[
h_{k-1}
\;=\;
\frac{2^{(k-1)+1} + (-1)^{k-1}}{3}
\;=\;
\frac{2^k + (-1)^{k-1}}{3}.
\]
We must show that the same formula then holds for \(h_k\).
From the recursive definition,
\[
h_k
\;=\; 2^k - h_{k-1}.
\]
Using the inductive hypothesis,
\[
h_k
\;=\; 2^k \;-\; \frac{2^k + (-1)^{k-1}}{3}
\;=\;
\frac{\,3 \cdot 2^k - \bigl(2^k + (-1)^{k-1}\bigr)}{3}
\;=\;
\frac{\,3 \cdot 2^k - 2^k - (-1)^{k-1}}{3}.
\]
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
\[
-\,(-1)^{k-1} \;=\; (-1)\cdot(-1)^{k-1} \;=\; (-1)^k.
\]
Hence
\[
2^{k+1} - (-1)^{k-1}
\;=\;
2^{k+1} + (-1)^k.
\]
Thus
\[
h_k
\;=\;
\frac{2^{k+1} + (-1)^k}{3},
\]
which is exactly the claimed closed-form formula.
