Answer
For every \(k \ge 2\),
\[
3\,a_{k-1} - 2\,a_{k-2} \;=\; a_k,
\]
which proves that any sequence of the form \(a_n = C\,2^n + D\) satisfies
\[
\boxed{a_k = 3\,a_{k-1} - 2\,a_{k-2}, \quad k \ge 2.}
\]
Work Step by Step
We want to show that for any real constants \(C\) and \(D\), the sequence defined by
\[
a_n \;=\; C\,2^n \;+\; D,
\quad \text{for all integers } n \ge 0,
\]
satisfies the second‐order linear recurrence
\[
a_k \;=\; 3\,a_{k-1} \;-\; 2\,a_{k-2},
\quad \text{for all integers } k \ge 2.
\]
---
## Proof
**Step 1. Write down the relevant terms.**
From the explicit formula:
\[
a_k = C\,2^k + D,\quad
a_{k-1} = C\,2^{k-1} + D,\quad
a_{k-2} = C\,2^{k-2} + D.
\]
---
**Step 2. Compute the right‐hand side of the recurrence.**
\[
3\,a_{k-1} - 2\,a_{k-2}
= 3\bigl(C\,2^{k-1} + D\bigr) \;-\; 2\bigl(C\,2^{k-2} + D\bigr).
\]
Distribute and group like terms:
\[
= 3C\,2^{k-1} \;+\; 3D
\;-\; 2C\,2^{k-2} \;-\; 2D
= \bigl(3C\,2^{k-1} \;-\; 2C\,2^{k-2}\bigr) \;+\; (3D - 2D).
\]
Simplify each group:
- Factor out \(2^{k-2}\) from the terms involving \(C\):
\[
3C\,2^{k-1} - 2C\,2^{k-2}
= 2^{k-2} \bigl(3C \cdot 2 - 2C\bigr)
= 2^{k-2} \bigl(6C - 2C\bigr)
= 2^{k-2} \cdot 4C
= 4C\,2^{k-2}
= C\,2^k.
\]
- For the \(D\) terms: \(3D - 2D = D\).
Thus,
\[
3\,a_{k-1} - 2\,a_{k-2}
= C\,2^k + D
= a_k.
\]
