Find `(x)/(1.2)+(x^(2))/(2.3)+(x^(3))/(3.4)+....=`

Options

A. `1+(1-x)/(x)log(1-x)`
B. `(1-x)/(x)log(1-x)`
C. `1+(1-x)/(x)log(1+x)`
D. `(1-x)/(x)log(1+x)`

Answer

To solve the series given by

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

we can express it in a more manageable form. Let's denote the general term of the series as

\[ a_n = \frac{x^n}{n(n+1)}. \]

### Step 1: Rewrite the General Term
We can rewrite the term \( \frac{1}{n(n+1)} \) using partial fractions:

\[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}. \]

Thus, we can express the general term as:

\[ a_n = x^n \left( \frac{1}{n} - \frac{1}{n+1} \right). \]

### Step 2: Sum the Series
Now, we can write the series \( S \) as:

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

This can be separated into two sums:

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

### Step 3: Evaluate Each Sum
The first sum is the Taylor series expansion for \( -\log(1-x) \):

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

For the second sum, we can shift the index:

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

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

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

Factoring out \( -\log(1-x) \):

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

### Final Result
Thus, the sum of the series is:

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