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 evaluate the infinite series given by:

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

This series can be expressed in summation notation as:

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

This series resembles the Taylor series expansion of the logarithmic function. We know that:

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

If we replace \(x\) with \(\frac{1}{x^2}\), we get:

\[ -\log\left(1 - \frac{1}{x^2}\right) = \frac{1}{x^2} + \frac{1}{2x^4} + \frac{1}{3x^6} + \ldots \]

Thus, we can write:

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

Next, we simplify \(1 - \frac{1}{x^2}\):

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

Now substituting this back into the logarithm:

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

Using the property of logarithms, we can separate the terms:

\[ S = -\left(\log(x^2 - 1) - \log(x^2)\right) = \log(x^2) - \log(x^2 - 1) \]

This can be further simplified:

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

Now, we know that \(y = 2x^2 - 1\). Therefore, we can express \(x^2\) in terms of \(y\):

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

Substituting this into our expression for \(S\):

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

This simplifies to:

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

Thus, the final result is:

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