Answer
1. For \(a_0 = 1\), \(a_1 = 3\):
\[
C = 2,\; D = -1,
\quad
a_2 = 7.
\]
2. For \(a_0 = 0\), \(a_1 = 2\):
\[
C = 2,\; D = -2,
\quad
a_2 = 6.
\]
Work Step by Step
We have an explicit formula for the sequence:
\[
a_n \;=\; C \cdot 2^n \;+\; D,
\quad n \ge 0,
\]
where \(C\) and \(D\) are real constants to be determined.
---
## (a) Find \(C\) and \(D\) so that \(a_0 = 1\) and \(a_1 = 3\). Then compute \(a_2\).
1. **Initial conditions**:
\[
a_0 = C \cdot 2^0 + D = C + D = 1,
\]
\[
a_1 = C \cdot 2^1 + D = 2C + D = 3.
\]
2. **Solve for \(C\) and \(D\)**:
Subtract the first equation from the second:
\[
(2C + D) - (C + D) = 3 - 1
\quad\Longrightarrow\quad
C = 2.
\]
Then from \(C + D = 1\), we have \(2 + D = 1\), so \(D = -1\).
3. **Hence**:
\[
C = 2,\quad D = -1,
\]
so
\[
a_n = 2 \cdot 2^n - 1 = 2^{n+1} - 1.
\]
4. **Compute \(a_2\)**:
\[
a_2 = 2^{2+1} - 1 = 2^3 - 1 = 8 - 1 = 7.
\]
---
## (b) Find \(C\) and \(D\) so that \(a_0 = 0\) and \(a_1 = 2\). Then compute \(a_2\).
1. **Initial conditions**:
\[
a_0 = C \cdot 2^0 + D = C + D = 0,
\]
\[
a_1 = C \cdot 2^1 + D = 2C + D = 2.
\]
2. **Solve for \(C\) and \(D\)**:
Subtract the first equation from the second:
\[
(2C + D) - (C + D) = 2 - 0
\quad\Longrightarrow\quad
C = 2.
\]
Then from \(C + D = 0\), we have \(2 + D = 0\), so \(D = -2\).
3. **Hence**:
\[
C = 2,\quad D = -2,
\]
so
\[
a_n = 2 \cdot 2^n - 2 = 2^{n+1} - 2.
\]
4. **Compute \(a_2\)**:
\[
a_2 = 2^{2+1} - 2 = 2^3 - 2 = 8 - 2 = 6.
\]
