# Any tricky method to solve this one?

The question : Prove that :$$\text{ if } y = 2x^2 - 1,\text{ then } \biggl[ \frac{1}{y} + \frac{1}{3y^3} + \frac{1}{5y^5}+ \cdots \biggr]$$ is equal to $$\frac{1}{2} \biggl[ \frac{1}{x^2} + \frac{1}{2x^4} + \frac{1}{3x^6} + \cdots \biggr]$$

Here I have modified the question, in the actual question (from my paper) there were four other options given,my approach was to reduce the first expression to $\frac{1}{2} \ln \biggl( \frac{y+1}{y-1} \biggr)$ and then trying to check for each options to find the match, now is there any other approach for this one ? Since it took me sometime for checking each options :( and the desired answer is at the last option!

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One you simplify to $\frac{1}{2}\log(\frac{y+1}{y-1})$, plug in $y = 2x^2 -1$ and simplify it to get $-\frac{1}{2} \log(1-\frac{1}{x^2})$. Now expand to get the desired answer. – user17762 Nov 22 '10 at 5:42
@Sivaram :I guess this is what I was missing!Thank you very much:) – Quixotic Nov 22 '10 at 5:45

One you simplify to $\frac{1}{2}\log(\frac{y+1}{y-1})$, plug in $y = 2x^2 -1$ and simplify it to get $-\frac{1}{2} \log(1-\frac{1}{x^2})$. Now expand to get the desired answer.