`sum_(n=1)^(oo)(1)/(2n-1)*x^(2n)=`

Options

A. a) `(x)/(2)log((1+x)/(1-x))^(x)`
B. b) `(1)/(2)log((1+x^(2))/(1-x^(2)))`
C. c) `x"log"(1+x)/(1-x)`
D. d) `(x)/(2)log((1+x)/(1-x))`

Answer

To solve the problem \( \sum_{n=1}^{\infty} \frac{1}{2n-1} x^{2n} \), we will follow these steps:

### Step 1: Rewrite the Series
We start with the series:
\[ S = \sum_{n=1}^{\infty} \frac{1}{2n-1} x^{2n} \]
This can be expressed as:
\[ S = x^2 + \frac{x^4}{3} + \frac{x^6}{5} + \cdots \]

### Step 2: Factor Out \( x^2 \)
Notice that each term in the series has a factor of \( x^2 \):
\[ S = x^2 \left( 1 + \frac{x^2}{3} + \frac{x^4}{5} + \cdots \right) \]

### Step 3: Recognize the Series as a Logarithmic Series
The series inside the parentheses can be related to the logarithmic series. Recall the series expansion for \( \log(1+x) \):
\[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \]
And for \( \log(1-x) \):
\[ \log(1-x) = -\left(x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots \right) \]

### Step 4: Combine the Logarithmic Series
Now, we can combine these two logarithmic series:
\[ \log(1+x) - \log(1-x) = \left( x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \right) - \left( -x - \frac{x^2}{2} - \frac{x^3}{3} - \cdots \right) \]
This simplifies to:
\[ \log(1+x) - \log(1-x) = 2x + \frac{2x^3}{3} + \frac{2x^5}{5} + \cdots \]

### Step 5: Relate to the Original Series
From the above, we can see that:
\[ \log(1+x) - \log(1-x) = 2 \left( x + \frac{x^3}{3} + \frac{x^5}{5} + \cdots \right) \]
Thus, we have:
\[ S = x^2 \cdot \frac{1}{2} \left( \log(1+x) - \log(1-x) \right) \]

### Step 6: Final Expression
Therefore, we can express \( S \) as:
\[ S = \frac{x^2}{2} \log\left(\frac{1+x}{1-x}\right) \]

### Conclusion
Thus, the final result for the series is:
\[ \sum_{n=1}^{\infty} \frac{1}{2n-1} x^{2n} = \frac{x}{2} \log\left(\frac{1+x}{1-x}\right) \]