If `|x|lt(1)/(2),` what is the value of
`1+[(x)/(1-x)]+[(n(n+1))/(2!)][(x)/(1-x)]^(2)+............+oo ?`

Options

A. `[(1-x)/(1-2x)]^(n)`
B. `(1-x)^(n)`
C. `[(1-2x)/(1-x)]^(n)`
D. `((1)/(1-x))^(n)`

Explanation

To solve the problem, we need to evaluate the infinite series given by:

\[ S = 1 + \frac{nx}{1-x} + \frac{n(n+1)}{2!} \left(\frac{x}{1-x}\right)^2 + \ldots \]

Given that \(|x| < \frac{1}{2}\), we can recognize that this series is a power series expansion.

### Step 1: Identify the series

The series can be recognized as the expansion of the function:

\[ S = \sum_{n=0}^{\infty} \frac{n(n-1)}{2!} \left(\frac{x}{1-x}\right)^n \]

This resembles the binomial series expansion.

### Step 2: Use the binomial series formula

The binomial series for \((1 - u)^{-k}\) is given by:

\[ (1 - u)^{-k} = \sum_{n=0}^{\infty} \frac{k(k+1)(k+2)\ldots(k+n-1)}{n!} u^n \]

In our case, we can rewrite the series as:

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

### Step 3: Simplify the expression

Now, we simplify \(\frac{x}{1-x}\):

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

We can simplify this further:

\[ 1 - \frac{x}{1-x} = \frac{1-x-x}{1-x} = \frac{1-2x}{1-x} \]

Thus,

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

### Step 4: Final result

The final result for the series is:

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

### Summary of Steps

1. Identify the series as a power series expansion.
2. Use the binomial series formula to express the series.
3. Simplify the expression to find the closed form.
4. State the final result.