Is there a way to express the reciprocal of the hypergeometric function 2F1(a,b;c;z) in terms of a b and c?

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I'm trying to use generating functions to get the value of some coefficients, namely

$\displaystyle\sum_{m\geq 0} f_{2m}x^m = 1 - \Big(\displaystyle\sum_{m\geq 0} u_{2m}x^m\Big)^{-1}\\$ and

$u_{2m} = \binom{2m}{m}4^{-2m}$ so that

$\displaystyle\sum_{m\geq 0} u_{2m}x^m = 2F1(\frac{1}{2},\frac{1}{2};1;x^2)$

(according to Wolfram Alpha).

For the first equation to work out and for me to find $f_{2m}$ I need to take $\displaystyle\sum_{m\geq 0} u_{2m}x^m$ out of the denominator. Any ideas or identities?