One way to do this is to recall that for [tex]|x|<1[/tex], we have
[tex]\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n[/tex]
so that
[tex]\displaystyle\frac x{10x^2+1}=\frac x{1-(-10x^2)}=x\sum_{n=0}^\infty(-10x^2)^n=\sum_{n=0}^\infty(-10)^nx^{2n+1}[/tex]
(which seems to match the first option) so long as [tex]|-10x^2|=10x^2<1[/tex], or [tex]-\frac1{\sqrt{10}}<x<\frac1{\sqrt{10}}[/tex], which is the interval of convergence.