Answer
See explanation
Work Step by Step
**Solution Explanation**
We have a sequence defined by
\[
d_n = 3^n \;-\; 2^n, \quad \text{for all integers } n \ge 0.
\]
We want to show it satisfies the recurrence relation
\[
d_k = 5\,d_{k-1} \;-\; 6\,d_{k-2},
\quad \text{for all integers } k \ge 2.
\]
---
### Step 1. Write down \(d_k\), \(d_{k-1}\), and \(d_{k-2}\)
From the explicit formula \(d_n = 3^n - 2^n\), we have:
\(d_k = 3^k - 2^k.\)
\(d_{k-1} = 3^{k-1} - 2^{k-1}.\)
\(d_{k-2} = 3^{k-2} - 2^{k-2}.\)
---
### Step 2. Verify the Recurrence
Compute the right-hand side of \(5\,d_{k-1} - 6\,d_{k-2}\):
\[
5\,d_{k-1} - 6\,d_{k-2}
\;=\;
5\bigl(3^{k-1} - 2^{k-1}\bigr)
\;-\;
6\bigl(3^{k-2} - 2^{k-2}\bigr).
\]
Separate terms involving powers of 3 and powers of 2:
\[
= \;5\,(3^{k-1}) \;-\; 5\,(2^{k-1})
\;-\; 6\,(3^{k-2}) \;+\; 6\,(2^{k-2}).
\]
1. **Combine the powers of 3**:
\[
5\,(3^{k-1}) \;-\; 6\,(3^{k-2})
= 3^{k-2}\,\bigl(5 \cdot 3 - 6\bigr)
= 3^{k-2}\,\bigl(15 - 6\bigr)
= 9\,(3^{k-2})
= 3^{k-2} \cdot 3^2
= 3^k.
\]
2. **Combine the powers of 2**:
\[
-5\,(2^{k-1}) \;+\; 6\,(2^{k-2})
= 2^{k-2}\,\bigl(-5 \cdot 2 + 6\bigr)
= 2^{k-2}\,\bigl(-10 + 6\bigr)
= -4\,(2^{k-2})
= -2^{k-2}\,\cdot 2^2
= -2^k.
\]
Putting these results together, we get
\[
5\,d_{k-1} - 6\,d_{k-2}
\;=\;
3^k \;-\; 2^k
\;=\;
d_k.
\]
Hence,
\[
d_k = 5\,d_{k-1} \;-\; 6\,d_{k-2},
\]
as required.
