Answer
**Any** sequence of the form \(b_n = C\,3^n + D\,(-2)^n\) satisfies the recurrence \(b_k = b_{k-1} + 6\,b_{k-2}\).
Work Step by Step
We want to show that for the sequence
\[
b_n \;=\; C \, 3^n \;+\; D \,\bigl(-2\bigr)^n,
\quad n \ge 0,
\]
(where \(C\) and \(D\) are arbitrary real constants), the terms satisfy the second‐order linear recurrence
\[
b_k \;=\; b_{k-1} \;+\; 6\,b_{k-2}
\quad\text{for all integers }k \ge 2.
\]
---
## Proof
1. **Write down the relevant terms** from the explicit formula:
\[
b_k = C\,3^k + D\,(-2)^k,
\quad
b_{k-1} = C\,3^{k-1} + D\,(-2)^{k-1},
\quad
b_{k-2} = C\,3^{k-2} + D\,(-2)^{k-2}.
\]
2. **Compute \(b_{k-1} + 6\,b_{k-2}\)** and simplify:
\[
b_{k-1} + 6\,b_{k-2}
\;=\; \bigl[C\,3^{k-1} + D\,(-2)^{k-1}\bigr]
\;+\; 6\,\bigl[C\,3^{k-2} + D\,(-2)^{k-2}\bigr].
\]
Group the \(C\)‐terms and the \(D\)‐terms:
\[
= C\bigl[3^{k-1} + 6\,3^{k-2}\bigr]
\;+\; D\bigl[(-2)^{k-1} + 6\,(-2)^{k-2}\bigr].
\]
Factor out common powers:
- For the \(C\)‐part:
\[
3^{k-1} = 3^{k-2} \cdot 3,
\quad
3^{k-1} + 6\,3^{k-2}
= 3^{k-2}\,\bigl(3 + 6\bigr)
= 9\,3^{k-2}
= 3^2\,3^{k-2}
= 3^k.
\]
- For the \(D\)‐part:
\[
(-2)^{k-1}
= (-2)^{k-2}\,\bigl(-2\bigr),
\quad
(-2)^{k-1} + 6\,(-2)^{k-2}
= (-2)^{k-2}\,\bigl[-2 + 6\bigr]
= 4\,(-2)^{k-2}
= (-2)^2 \,(-2)^{k-2}
= (-2)^k.
\]
Putting these back in, we get
\[
b_{k-1} + 6\,b_{k-2}
= C\,3^k + D\,(-2)^k
= b_k.
\]
Hence,
\[
\boxed{b_k \;=\; b_{k-1} + 6\,b_{k-2},\quad \text{for all } k \ge 2.}
\]
