To find the sum to n terms of the given series, we will analyze each series separately and apply the formula for the sum of an arithmetic-geometric progression.
### (i) Series: \(1 + \frac{3}{2} + \frac{5}{2^2} + \frac{7}{2^3} + \ldots\)
1. **Identify the first term (a)**:
- The first term \(a = 1\).
2. **Identify the common difference (d)**:
- The numerators form an arithmetic progression: \(1, 3, 5, 7, \ldots\)
- The common difference \(d = 2\).
3. **Identify the common ratio (r)**:
- The denominators form a geometric progression: \(1, 2, 2^2, 2^3, \ldots\)
- The common ratio \(r = \frac{1}{2}\).
4. **Apply the formula for the sum of n terms (S_n)**:
\[
S_n = \frac{a}{1 - r} + \frac{d \cdot r}{(1 - r^2)} \cdot (1 - r^{n-1})
\]
Plugging in the values:
\[
S_n = \frac{1}{1 - \frac{1}{2}} + \frac{2 \cdot \frac{1}{2}}{(1 - \left(\frac{1}{2}\right)^2)} \cdot (1 - \left(\frac{1}{2}\right)^{n-1})
\]
5. **Simplify the expression**:
- The first term simplifies to:
\[
\frac{1}{\frac{1}{2}} = 2
\]
- The second term simplifies as follows:
\[
\frac{2 \cdot \frac{1}{2}}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}
\]
Thus, we have:
\[
S_n = 2 + \frac{4}{3} \cdot (1 - \frac{1}{2^{n-1}})
\]
6. **Final expression for S_n**:
\[
S_n = 2 + \frac{4}{3} - \frac{4}{3 \cdot 2^{n-1}} = \frac{6}{3} + \frac{4}{3} - \frac{4}{3 \cdot 2^{n-1}} = \frac{10}{3} - \frac{4}{3 \cdot 2^{n-1}}
\]
### (ii) Series: \(1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \ldots\)
1. **Identify the first term (a)**:
- The first term \(a = 1\).
2. **Identify the common difference (d)**:
- The numerators form an arithmetic progression: \(1, 2, 3, 4, \ldots\)
- The common difference \(d = 1\).
3. **Identify the common ratio (r)**:
- The denominators form a geometric progression: \(1, 3, 3^2, 3^3, \ldots\)
- The common ratio \(r = \frac{1}{3}\).
4. **Apply the formula for the sum of n terms (S_n)**:
\[
S_n = \frac{1}{1 - \frac{1}{3}} + \frac{1 \cdot \frac{1}{3}}{(1 - \left(\frac{1}{3}\right)^2)} \cdot (1 - \left(\frac{1}{3}\right)^{n-1})
\]
Plugging in the values:
\[
S_n = \frac{1}{\frac{2}{3}} + \frac{\frac{1}{3}}{(1 - \frac{1}{9})} \cdot (1 - \left(\frac{1}{3}\right)^{n-1})
\]
5. **Simplify the expression**:
- The first term simplifies to:
\[
\frac{3}{2}
\]
- The second term simplifies as follows:
\[
\frac{\frac{1}{3}}{\frac{8}{9}} = \frac{3}{8}
\]
Thus, we have:
\[
S_n = \frac{3}{2} + \frac{3}{8} \cdot (1 - \frac{1}{3^{n-1}})
\]
6. **Final expression for S_n**:
\[
S_n = \frac{3}{2} + \frac{3}{8} - \frac{3}{8 \cdot 3^{n-1}} = \frac{12}{8} + \frac{3}{8} - \frac{3}{8 \cdot 3^{n-1}} = \frac{15}{8} - \frac{3}{8 \cdot 3^{n-1}}
\]
### Summary of Results:
- For series (i):
\[
S_n = \frac{10}{3} - \frac{4}{3 \cdot 2^{n-1}}
\]
- For series (ii):
\[
S_n = \frac{15}{8} - \frac{3}{8 \cdot 3^{n-1}}
\]
