[tex]\dfrac{x^2}{2-x^2}=\dfrac{x^2}2\cdot\dfrac1{1-\frac{x^2}2}[/tex]
Since
[tex]\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n[/tex]
for [tex]|x|<1|[/tex], the power series for the rightmost rational expression is
[tex]\dfrac1{1-\frac{x^2}2}=\displaystyle\sum_{n=0}^\infty\left(\frac{x^2}2\right)^n=\sum_{n=0}^\infty \frac{x^{2n}}{2^n}[/tex]
which is applicable for [tex]\left|\dfrac{x^2}2\right|<1[/tex], or [tex]|x|<\sqrt2[/tex].
Distributing the other rational expression gives
[tex]\dfrac{x^2}{2-x^2}=\displaystyle\frac{x^2}2\sum_{n=0}^\infty\frac{x^{2n}}{2^n}=\sum_{n=0}^\infty\frac{x^{2n+2}}{2^{n+1}}[/tex]
I'm not sure what is meant by "simplify the summation to a single variable", but you can rewrite this as
[tex]\displaystyle\sum_{n=0}^\infty\frac{x^{2(n+1)}}{2^{n+1}}=\sum_{n=0}^\infty\left(\frac{x^2}2\right)^{n+1}[/tex]
which could be interpreted as a simpler form, but that's really in the eye of the beholder...
