Find the sum to infinity of the following series:
(i) `2+3/2+1+5/8+`…………
(ii) `1+2/3+3/(3^(2))+4/(3^(3))+`………..
(iii) `1-3/2+5/4-7/8+`……..
(iv) `1-2/3+3/(3^(2))-4/(3^(3))+5/(3^(4))-`………..

Answer

To find the sum to infinity of the given series, we will use the formula for the sum of an Arithmetic-Geometric Progression (AGP):

\[ S_{\infty} = \frac{a \cdot b}{1 - r} + \frac{d \cdot b \cdot r}{(1 - r)^2} \]

where:
- \( a \) is the first term of the arithmetic sequence,
- \( d \) is the common difference of the arithmetic sequence,
- \( b \) is the first term of the geometric sequence,
- \( r \) is the common ratio of the geometric sequence.

### (i) Series: \( 2 + \frac{3}{2} + 1 + \frac{5}{8} + \ldots \)

1. Identify the arithmetic sequence:
- First term \( a = 2 \)
- Common difference \( d = 1 - \frac{3}{2} = -\frac{1}{2} \)

2. Identify the geometric sequence:
- First term \( b = 1 \)
- Common ratio \( r = \frac{1}{2} \)

3. Substitute into the formula:
\[ S_{\infty} = \frac{2 \cdot 1}{1 - \frac{1}{2}} + \frac{-\frac{1}{2} \cdot 1 \cdot \frac{1}{2}}{(1 - \frac{1}{2})^2} \]
\[ = \frac{2}{\frac{1}{2}} - \frac{\frac{1}{4}}{\frac{1}{4}} = 4 - 1 = 3 \]

### (ii) Series: \( 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \ldots \)

1. Identify the arithmetic sequence:
- First term \( a = 1 \)
- Common difference \( d = 1 \)

2. Identify the geometric sequence:
- First term \( b = 1 \)
- Common ratio \( r = \frac{1}{3} \)

3. Substitute into the formula:
\[ S_{\infty} = \frac{1 \cdot 1}{1 - \frac{1}{3}} + \frac{1 \cdot 1 \cdot \frac{1}{3}}{(1 - \frac{1}{3})^2} \]
\[ = \frac{1}{\frac{2}{3}} + \frac{\frac{1}{3}}{\left(\frac{2}{3}\right)^2} = \frac{3}{2} + \frac{\frac{1}{3}}{\frac{4}{9}} = \frac{3}{2} + \frac{3}{4} = \frac{6}{4} + \frac{3}{4} = \frac{9}{4} \]

### (iii) Series: \( 1 - \frac{3}{2} + \frac{5}{4} - \frac{7}{8} + \ldots \)

1. Identify the arithmetic sequence:
- First term \( a = 1 \)
- Common difference \( d = 2 \)

2. Identify the geometric sequence:
- First term \( b = 1 \)
- Common ratio \( r = -\frac{1}{2} \)

3. Substitute into the formula:
\[ S_{\infty} = \frac{1 \cdot 1}{1 - (-\frac{1}{2})} + \frac{2 \cdot 1 \cdot (-\frac{1}{2})}{(1 - (-\frac{1}{2}))^2} \]
\[ = \frac{1}{\frac{3}{2}} + \frac{-1}{\left(\frac{3}{2}\right)^2} = \frac{2}{3} - \frac{-1}{\frac{9}{4}} = \frac{2}{3} + \frac{4}{9} \]
\[ = \frac{6}{9} + \frac{4}{9} = \frac{10}{9} \]

### (iv) Series: \( 1 - \frac{2}{3} + \frac{3}{3^2} - \frac{4}{3^3} + \ldots \)

1. Identify the arithmetic sequence:
- First term \( a = 1 \)
- Common difference \( d = 1 \)

2. Identify the geometric sequence:
- First term \( b = 1 \)
- Common ratio \( r = -\frac{1}{3} \)

3. Substitute into the formula:
\[ S_{\infty} = \frac{1 \cdot 1}{1 - (-\frac{1}{3})} + \frac{1 \cdot 1 \cdot (-\frac{1}{3})}{(1 - (-\frac{1}{3}))^2} \]
\[ = \frac{1}{\frac{4}{3}} + \frac{-\frac{1}{3}}{\left(\frac{4}{3}\right)^2} = \frac{3}{4} - \frac{-\frac{1}{3}}{\frac{16}{9}} = \frac{3}{4} + \frac{3}{16} \]
\[ = \frac{12}{16} + \frac{3}{16} = \frac{15}{16} \]

### Summary of Results:
1. \( S_{\infty} \) for (i) = 3
2. \( S_{\infty} \) for (ii) = \( \frac{9}{4} \)
3. \( S_{\infty} \) for (iii) = \( \frac{10}{9} \)
4. \( S_{\infty} \) for (iv) = \( \frac{15}{16} \)