Answer
\[
\boxed{d_k = 5 \cdot 2^{k-1} \;-\; 3.}
\]
Work Step by Step
We have the recurrence
\[
d_k = 2 d_{k-1} + 3 \quad \text{for } k \ge 2,
\quad\text{with}\quad d_1 = 2.
\]
A standard way to solve this kind of first-order linear recurrence is:
1. **Solve the homogeneous part**: \(d_k^{(h)} = 2 \, d_{k-1}^{(h)}.\)
The general solution to this homogeneous recurrence is
\[
d_k^{(h)} = C \cdot 2^{k-1},
\]
for some constant \(C\).
2. **Find a particular solution**: Guess that \(d_k^{(p)}\) is a constant, say \(A\). Plugging into the original recurrence:
\[
A = 2A + 3 \quad \Longrightarrow \quad -A = 3 \quad \Longrightarrow \quad A = -3.
\]
So a particular solution is \(d_k^{(p)} = -3\).
3. **Combine to get the general solution**:
\[
d_k = d_k^{(h)} + d_k^{(p)} = C \cdot 2^{k-1} - 3.
\]
4. **Use the initial condition** \(d_1 = 2\) to solve for \(C\). When \(k=1\),
\[
d_1 = C \cdot 2^{1-1} - 3 = C - 3.
\]
Since \(d_1 = 2\), we get \(C - 3 = 2 \implies C = 5\).
