Find the sum of
`x+2x^(2)+3x^(3)+4x^(4)`+...

Answer

To find the sum of the series \( S = x + 2x^2 + 3x^3 + 4x^4 + \ldots \) up to infinity, we can follow these steps:

### Step-by-Step Solution:

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

2. **Multiply the Series by \( x \)**:
Multiply both sides of the equation by \( x \):
\[ xS = x^2 + 2x^3 + 3x^4 + 4x^5 + \ldots \]

3. **Align the Series**:
Now, align the two equations:
\[ S = x + 2x^2 + 3x^3 + 4x^4 + \ldots \quad \text{(Equation 1)} \]
\[ xS = x^2 + 2x^3 + 3x^4 + 4x^5 + \ldots \quad \text{(Equation 2)} \]

4. **Subtract Equation 2 from Equation 1**:
Subtract Equation 2 from Equation 1:
\[ S - xS = (x + 2x^2 + 3x^3 + 4x^4 + \ldots) - (x^2 + 2x^3 + 3x^4 + 4x^5 + \ldots) \]

5. **Simplify the Left Side**:
Factor out \( S \) on the left side:
\[ S(1 - x) = x + (2x^2 - x^2) + (3x^3 - 2x^3) + (4x^4 - 3x^4) + \ldots \]
This simplifies to:
\[ S(1 - x) = x + x^2 + x^3 + x^4 + \ldots \]

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

7. **Substitute Back**:
Substitute the sum of the geometric series back into the equation:
\[ S(1 - x) = \frac{x}{1 - x} \]

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

### Final Answer:
Thus, the sum of the series \( S = x + 2x^2 + 3x^3 + 4x^4 + \ldots \) is:
\[ S = \frac{x}{(1 - x)^2} \quad \text{for } |x| < 1 \]

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