If `f(x)=(x^(2))/(1.2)-(x^(3))/(2.3)+(x^(4))/(3.4)-(x^(5))/(4.5)+..oo` then

Options

A. a) `log_(e )((1+x)/(1-x))`
B. b) `log_(e )(1+x) `
C. c) `(1+x)log_(e )(1+x)`- x
D. d) `log_(e )(1-x)`

Answer

To solve the problem, we need to analyze the function given:

\[ f(x) = \frac{x^2}{1 \cdot 2} - \frac{x^3}{2 \cdot 3} + \frac{x^4}{3 \cdot 4} - \frac{x^5}{4 \cdot 5} + \ldots \]

### Step 1: Recognize the Series
The series can be recognized as a power series. The general term appears to be:

\[ (-1)^{n} \frac{x^{n+1}}{n(n+1)} \]

for \( n = 1, 2, 3, \ldots \)

### Step 2: Relate to Logarithmic Functions
We know the Taylor series expansions for logarithmic functions:

1. For \( \log(1+x) \):
\[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]

2. For \( \log(1-x) \):
\[ \log(1-x) = -\left( x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \ldots \right) \]

### Step 3: Combine the Logarithmic Functions
We can combine these two logarithmic functions:

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

This results in:

\[ \log(1+x) - \log(1-x) = 2\left(x + \frac{x^3}{3} + \frac{x^5}{5} + \ldots\right) \]

### Step 4: Find the Expression for \( f(x) \)
From the above, we can see that:

\[ f(x) = 1 + x \log(1+x) - x \]

This means we can express \( f(x) \) as:

\[ f(x) = 1 + x \log(1+x) - x \]

### Step 5: Verify the Result
To verify, we can differentiate \( f(x) \) or check the series expansion of \( 1 + x \log(1+x) - x \) to ensure it matches the original series.

### Final Result
Thus, we conclude that:

\[ f(x) = 1 + x \log(1+x) - x \]