The sum of the series
` (1)/(2)x^(2) + (2)/(3)x^(3) + (3)/(4)x^(4) + (4)/(5) x^(5) +...` equal to

Options

A. `(x)/(1+x) + log (1+x)`
B. `(x)/(1-x) + log(1-x)`
C. `-(x)/(1+x) + log (1+x)`
D. none of these

Answer

To find the sum of the series

\[ S = \frac{1}{2}x^2 + \frac{2}{3}x^3 + \frac{3}{4}x^4 + \frac{4}{5}x^5 + \ldots \]

we will follow a systematic approach.

### Step 1: Rewrite the series in a more manageable form
We can express the series in terms of a general term:

\[ S = \sum_{n=2}^{\infty} \frac{n-1}{n} x^n \]

This can be rewritten as:

\[ S = \sum_{n=2}^{\infty} \left(1 - \frac{1}{n}\right) x^n \]

### Step 2: Split the series
Now, we can separate the series into two parts:

\[ S = \sum_{n=2}^{\infty} x^n - \sum_{n=2}^{\infty} \frac{1}{n} x^n \]

### Step 3: Calculate the first sum
The first sum is a geometric series:

\[ \sum_{n=2}^{\infty} x^n = x^2 + x^3 + x^4 + \ldots = \frac{x^2}{1-x} \quad \text{(for } |x| < 1\text{)} \]

### Step 4: Calculate the second sum
The second sum can be recognized as a series related to the logarithm:

\[ \sum_{n=2}^{\infty} \frac{1}{n} x^n = x^2 \sum_{n=0}^{\infty} \frac{x^n}{n+2} = x^2 \left( -\log(1-x) - x \right) \quad \text{(for } |x| < 1\text{)} \]

### Step 5: Combine the results
Now we substitute back into our expression for \( S \):

\[ S = \frac{x^2}{1-x} - \left( -\log(1-x) - x \right) \]

This simplifies to:

\[ S = \frac{x^2}{1-x} + x + \log(1-x) \]

### Step 6: Final expression
Thus, we can write:

\[ S = \frac{x}{1-x} + \log(1-x) \]

### Final Answer:
The sum of the series is:

\[ S = \frac{x}{1-x} + \log(1-x) \]