(i) `1+2x+3x^(2)+4x^(3)I+`………….. When `|x|lt1`
(ii) `1+3x+5x^(2)+7x^(3)+`…………. When `|x|lt1` (iii) `1+4x+7x^(2)+10x^(3)+`……………..when`|x|lt`1.

Answer

To solve the given series, we will approach each part step by step.

### Part (i): Series `1 + 2x + 3x^2 + 4x^3 + ...`

1. **Define the series**: Let \( S = 1 + 2x + 3x^2 + 4x^3 + \ldots \)

2. **Multiply by \( x \)**: Multiply the entire series by \( x \):
\[ Sx = 2x + 3x^2 + 4x^3 + 5x^4 + \ldots \]

3. **Subtract the two equations**: Now, subtract \( Sx \) from \( S \):
\[ S - Sx = 1 + (2x - 2x) + (3x^2 - 3x^2) + (4x^3 - 4x^3) + \ldots \]
This simplifies to:
\[ S(1 - x) = 1 + x + x^2 + x^3 + \ldots \]

4. **Recognize the geometric series**: The right-hand side is a geometric series with first term 1 and common ratio \( x \):
\[ 1 + x + x^2 + x^3 + \ldots = \frac{1}{1 - x} \quad \text{(for } |x| < 1\text{)} \]

5. **Substitute back**: Now we have:
\[ S(1 - x) = \frac{1}{1 - x} \]

6. **Solve for \( S \)**:
\[ S = \frac{1}{(1 - x)^2} \]

### Part (ii): Series `1 + 3x + 5x^2 + 7x^3 + ...`

1. **Define the series**: Let \( S = 1 + 3x + 5x^2 + 7x^3 + \ldots \)

2. **Identify the general term**: The general term can be expressed as \( (2n - 1)x^{n-1} \).

3. **Multiply by \( x \)**:
\[ Sx = 3x + 5x^2 + 7x^3 + \ldots \]

4. **Subtract**:
\[ S - Sx = 1 + (3x - 2x) + (5x^2 - 3x^2) + (7x^3 - 5x^3) + \ldots \]
This simplifies to:
\[ S(1 - x) = 1 + x + 2x^2 + 3x^3 + \ldots \]

5. **Recognize the new series**: The right-hand side can be expressed as:
\[ 1 + x + 2x^2 + 3x^3 + \ldots = \frac{1 + x}{(1 - x)^2} \]

6. **Substitute back**:
\[ S(1 - x) = \frac{1 + x}{(1 - x)^2} \]

7. **Solve for \( S \)**:
\[ S = \frac{1 + x}{(1 - x)^3} \]

### Part (iii): Series `1 + 4x + 7x^2 + 10x^3 + ...`

1. **Define the series**: Let \( S = 1 + 4x + 7x^2 + 10x^3 + \ldots \)

2. **Identify the general term**: The general term can be expressed as \( (3n - 2)x^{n-1} \).

3. **Multiply by \( x \)**:
\[ Sx = 4x + 7x^2 + 10x^3 + \ldots \]

4. **Subtract**:
\[ S - Sx = 1 + (4x - 3x) + (7x^2 - 4x^2) + (10x^3 - 7x^3) + \ldots \]
This simplifies to:
\[ S(1 - x) = 1 + x + 3x^2 + 6x^3 + \ldots \]

5. **Recognize the new series**: The right-hand side can be expressed as:
\[ 1 + x + 3x^2 + 6x^3 + \ldots = \frac{1 + 3x}{(1 - x)^3} \]

6. **Substitute back**:
\[ S(1 - x) = \frac{1 + 3x}{(1 - x)^3} \]

7. **Solve for \( S \)**:
\[ S = \frac{1 + 3x}{(1 - x)^4} \]

### Final Answers:
1. \( S = \frac{1}{(1 - x)^2} \) for part (i)
2. \( S = \frac{1 + x}{(1 - x)^3} \) for part (ii)
3. \( S = \frac{1 + 3x}{(1 - x)^4} \) for part (iii)