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

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 can express the series in a more manageable form.

### Step 1: Rewrite the series
We can rewrite the series as:

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

This can be split into two parts:

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

### Step 2: Calculate the first sum
The first sum is a geometric series starting from \(n=2\):

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

### Step 3: Calculate the second sum
The second sum can be related to the Taylor series expansion of \(-\log(1-x)\):

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

Thus,

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

### Step 4: Combine the results
Now substituting back into the expression for \(S\):

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

This simplifies to:

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

### Step 5: Final expression
Thus, the sum of the series is:

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

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

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