If\[y=2{{x}^{2}}-1\], then\[\left[ \frac{1}{y}+\frac{1}{3{{y}^{3}}}+\frac{1}{5{{y}^{5}}}+.... \right]\] is equal to

\[\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}+\frac{1}{3{{x}^{6}}}+\frac{1}{5{{x}^{10}}}+..... \right]\]

\[\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}-\frac{1}{2{{x}^{4}}}+\frac{1}{3{{x}^{6}}}-..... \right]\]

\[\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}+\frac{1}{2{{x}^{4}}}+\frac{1}{3{{x}^{6}}}+..... \right]\]

\[\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}-\frac{1}{3{{x}^{6}}}+\frac{1}{5{{x}^{10}}}-..... \right]\]

If\[y=2{{x}^{2}}-1\], then\[\left[ \frac{1}{y}+\frac{1}{3{{y}^{3}}}+\f

1.	If\[y=2{{x}^{2}}-1\], then\[\left[ \frac{1}{y}+\frac{1}{3{{y}^{3}}}+\frac{1}{5{{y}^{5}}}+.... \right]\] is equal to
A.	\[\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}-\frac{1}{2{{x}^{4}}}+\frac{1}{3{{x}^{6}}}-..... \right]\]
B.	\[\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}+\frac{1}{2{{x}^{4}}}+\frac{1}{3{{x}^{6}}}+..... \right]\]
C.	\[\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}+\frac{1}{3{{x}^{6}}}+\frac{1}{5{{x}^{10}}}+..... \right]\]
D.	\[\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}-\frac{1}{3{{x}^{6}}}+\frac{1}{5{{x}^{10}}}-..... \right]\]
Answer» C. \[\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}+\frac{1}{3{{x}^{6}}}+\frac{1}{5{{x}^{10}}}+..... \right]\]