Answer
1. **(a)** \(b_0 = 0, b_1 = 5\):
\[
C = 1,\; D = -1,
\quad
\text{so } b_n = 3^n - (-2)^n,
\quad
b_2 = 5.
\]
2. **(b)** \(b_0 = 3, b_1 = 4\):
\[
C = 2,\; D = 1,
\quad
\text{so } b_n = 2\cdot 3^n + (-2)^n,
\quad
b_2 = 22.
\]
Work Step by Step
We have a sequence \(\{b_n\}\) defined by the explicit formula
\[
b_n \;=\; C \cdot 3^n \;+\; D \cdot (-2)^n,
\quad\text{for all integers } n \ge 0,
\]
where \(C\) and \(D\) are real constants to be determined.
---
## (a) Find \(C\) and \(D\) so that \(b_0 = 0\) and \(b_1 = 5\). Then compute \(b_2\).
1. **Initial conditions**:
\[
b_0 = C \cdot 3^0 + D \cdot (-2)^0 = C + D = 0,
\]
\[
b_1 = C \cdot 3^1 + D \cdot (-2)^1 = 3C \;+\; (-2)D = 5.
\]
2. **Solve for \(C\) and \(D\)**:
From \(b_0 = 0\), we get
\[
C + D = 0
\quad\Longrightarrow\quad
D = -C.
\]
Substitute \(D = -C\) into \(b_1 = 5\):
\[
3C + (-2)\bigl(-C\bigr) = 3C + 2C = 5C = 5,
\]
hence \(C = 1\). Then \(D = -1\).
3. **Thus**:
\[
C = 1,\quad D = -1,
\]
so
\[
b_n = 3^n \;-\; (-2)^n.
\]
4. **Compute \(b_2\)**:
\[
b_2
= 3^2 \;-\; (-2)^2
= 9 \;-\; 4
= 5.
\]
---
## (b) Find \(C\) and \(D\) so that \(b_0 = 3\) and \(b_1 = 4\). Then compute \(b_2\).
1. **Initial conditions**:
\[
b_0 = C \cdot 3^0 + D \cdot (-2)^0 = C + D = 3,
\]
\[
b_1 = C \cdot 3^1 + D \cdot (-2)^1 = 3C + (-2)D = 4.
\]
2. **Solve for \(C\) and \(D\)**:
From \(b_0 = 3\),
\[
C + D = 3
\quad\Longrightarrow\quad
D = 3 - C.
\]
Substitute into \(b_1 = 4\):
\[
3C + (-2)\bigl(3 - C\bigr)
= 3C - 6 + 2C
= 5C - 6
= 4.
\]
Hence \(5C = 10\), so \(C = 2\). Then \(D = 3 - 2 = 1\).
3. **Thus**:
\[
C = 2,\quad D = 1,
\]
so
\[
b_n = 2 \cdot 3^n \;+\; 1 \cdot (-2)^n.
\]
4. **Compute \(b_2\)**:
\[
b_2
= 2 \cdot 3^2 + 1 \cdot (-2)^2
= 2 \cdot 9 + 4
= 18 + 4
= 22.
\]
