That's a large sequence!

This is the sigma notation-

$\sum _{ n=1 }^{ \infty }({ \frac { { x }^{ 2n-2 } }{ (1-{ x }^{ 2n+1 })(1-{ x }^{ 2n-1 }) } })$

$\frac{1}{{x}^{2}}\sum_{n=1}^{\infty}({\frac{{x}^{2n}}{(1-{x}^{2n+1})(1-{x}^{2n-1})}})$

Now we can split the term in the following way -

$1 - {x}^{2n+1} - 1 + {x}^{2n-1} = {x}^{2n}(\frac{1}{x} - x)$

$\frac{1}{{x}^{2}}\sum_{n=1}^{\infty}({\frac{1}{-x+\frac{1}{x}}\frac{1}{1-{x}^{2n-1}} - \frac{1}{-x+\frac{1}{x}}\frac{1}{1-{x}^{2n+1}}})$

$\frac{1}{x(1-{x}^{2})}\sum_{n=1}^{\infty}({\frac{1}{1-{x}^{2n-1}} - \frac{1}{1-{x}^{2n+1}}})$

Now, this is a telescoping series converging to $\frac{1}{1-x}$

And hence, the total sum converges to -

$\frac{1}{x(1-{x}^{2})(1-x)}$

$\frac{1}{x({(1-x)}^{2})(1+x)}$

NOT A SOLID SOLUTION: I've just checked first term for $x=\frac { 1 }{ 2 }$ , it comes out to be $\frac { 16 }{ 7 }$ , hence, options A,C are rejected, also D becomes 8 for $x=\frac { 1 }{ 2 }$ , which is a very large value for this converging series.

$\dfrac{1}{(1-x)^2}$ is a common factor and all quantities are + tive. |x|<1 so its higher powers approach zero. So $\dfrac{1}{(1+x)*(1-x)^2}$ can only be the solution.