Let $A\in R^{n\times n}$ (invertible and symmetric), $B\in R^{k\times k}$ (invertible and symmetric), $C\in R^{n\times k}$, $D\in R^{k\times n}$, $CBD$ is (invertible and symmetric). I am trying to obtain an expression for $(A-CBD)^{-1}$, which does not involve the inverses of $A$ and $CBD$.
As a consequence of the Woodbury identity (or binomial theorem), $$ (A-CBD)^{-1}=\sum_{j=0}^{\infty}[A^{-1} (CBD)]^j \times A^{-1}, $$ but this involves $A^{-1}$, which is unknown. Any suggestions?