Sum up to n terms the series
1+ 2x + `3x^(2) + 4x^(3) + ... `

Answer

To find the sum of the series \( S_n = 1 + 2x + 3x^2 + 4x^3 + \ldots + nx^{n-1} \), we can use a systematic approach. Here’s a step-by-step solution:

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

### Step 2: Multiply the Series by \( x \)
Now, multiply the entire series \( S \) by \( x \):
\[ xS = x + 2x^2 + 3x^3 + 4x^4 + \ldots \]

### Step 3: Subtract the Two Equations
Now, subtract the second equation from the first:
\[ S - xS = (1 + 2x + 3x^2 + 4x^3 + \ldots) - (x + 2x^2 + 3x^3 + 4x^4 + \ldots) \]
This simplifies to:
\[ S - xS = 1 + (2x - x) + (3x^2 - 2x^2) + (4x^3 - 3x^3) + \ldots \]
\[ S - xS = 1 + x + x^2 + x^3 + \ldots \]

### Step 4: Recognize the Right Side as a Geometric Series
The right-hand side \( 1 + x + x^2 + x^3 + \ldots \) is a geometric series with first term \( 1 \) and common ratio \( x \). The sum of an infinite geometric series is given by:
\[ \text{Sum} = \frac{1}{1 - x} \quad \text{(for } |x| < 1\text{)} \]
Thus,
\[ S - xS = \frac{1}{1 - x} \]

### Step 5: Factor Out \( S \)
Now, factor out \( S \) from the left side:
\[ S(1 - x) = \frac{1}{1 - x} \]

### Step 6: Solve for \( S \)
Now, solve for \( S \):
\[ S = \frac{1}{(1 - x)^2} \]

### Final Result
Thus, the sum of the series up to \( n \) terms is:
\[ S_n = \frac{1}{(1 - x)^2} \]