Answer
See explanation
Work Step by Step
Below is a succinct inductive proof that the sequence defined by
\[
d_k \;=\; 2\,d_{k-1} \;+\; 3
\quad \text{for } k \ge 2,
\quad \text{with} \quad d_1 = 2,
\]
is given explicitly by
\[
\boxed{d_k = 5 \cdot 2^{\,k-1} \;-\; 3}.
\]
---
## **1. Base Case** \((k = 1)\)
We must check that the closed‐form formula holds for \(k=1\). The given initial condition is
\[
d_1 = 2.
\]
On the other hand, plugging \(k=1\) into the proposed formula,
\[
5 \cdot 2^{\,1-1} - 3
= 5 \cdot 2^0 - 3
= 5 \cdot 1 - 3
= 2.
\]
Hence \(d_1 = 2\) matches the formula, and the base case is verified.
---
## **2. Inductive Step**
Suppose for some integer \(k \ge 2\) the formula holds at \(k-1\), that is,
\[
d_{k-1} \;=\; 5 \cdot 2^{\,(k-1)-1} \;-\; 3
\;=\;
5 \cdot 2^{\,k-2} \;-\; 3.
\]
We must show that it then holds at \(k\). By the recurrence relation,
\[
d_k
\;=\; 2\,d_{k-1} + 3.
\]
Using the inductive hypothesis for \(d_{k-1}\), we get
\[
d_k
= 2 \Bigl[5 \cdot 2^{\,k-2} - 3\Bigr] + 3
= 2 \cdot 5 \cdot 2^{\,k-2} - 2 \cdot 3 + 3
= 5 \cdot 2^{\,k-1} - 6 + 3
= 5 \cdot 2^{\,k-1} - 3.
\]
This is exactly the closed‐form we wanted to prove for \(d_k\).
