If `y=2x^(2)-1` then `(1)/(x^(2))+(1)/(2x^(4))+(1)/(3x^(6))+…infty` equals to

Options

A. -`log_(e)((y-1)/(y+1))`
B. `log_(e)((1+y)/(1-y))`
C. `log_(e)((1-y)/(1+y))`
D. `log((1+2y)/(1-2y))`

Answer

To solve the problem, we need to find the sum of the series:

\[ \frac{1}{x^2} + \frac{1}{2x^4} + \frac{1}{3x^6} + \ldots \]

This series can be expressed in a more manageable form. We can rewrite it as:

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

This is a standard series that can be related to the logarithmic function. The series:

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

can be differentiated to find a series that resembles our original series.

### Step 1: Recognize the series
We can recognize that our series is similar to the Taylor series expansion of \(-\ln(1-x)\). Specifically, we can relate it to:

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

### Step 2: Substitute \(x\) with \(\frac{1}{x^2}\)
To match our series, we substitute \(x\) with \(\frac{1}{x^2}\):

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

### Step 3: Simplify the logarithm
Now we can simplify the logarithm:

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

### Step 4: Use the given expression for \(y\)
From the problem, we know that \(y = 2x^2 - 1\). We can express \(x^2\) in terms of \(y\):

\[ x^2 = \frac{y + 1}{2} \]

### Step 5: Substitute \(x^2\) into the logarithm
Now we substitute \(x^2\) into our logarithmic expression:

\[ -\ln\left(\frac{y + 1}{2} - 1\right) + \ln\left(\frac{y + 1}{2}\right) \]

This simplifies to:

\[ -\ln\left(\frac{y - 1}{2}\right) + \ln\left(\frac{y + 1}{2}\right) \]

### Step 6: Combine the logarithms
Using the properties of logarithms, we can combine these:

\[ \ln\left(\frac{y + 1}{2}\right) - \ln\left(\frac{y - 1}{2}\right) = \ln\left(\frac{y + 1}{y - 1}\right) \]

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

\[ \ln\left(\frac{y + 1}{y - 1}\right) \]