Let $G$ and $G+E$ be invertible matrices where $E$ is a rank-$1$ matrix. Let $g=\mathrm{tr}(EG^{-1})$. The Sherman-Morrison formula is $$ (G+E)^{-1}=G^{-1}-\frac{1}{1+g}G^{-1}EG^{-1}.$$ In Miller (1981), the author discusses the inverse of $G+H$ where $G$ and $G+H$ are invertible and $H$ has positive rank $r$. A recursive formula is obtained after defining rank-$1$ matrices $E_i$'s such that $H=E_1+E_2+\cdots+E_r$: $$C_{k+1}^{-1}=C_{k}^{-1}-v_kC_{k}^{-1}E_kC_{k}^{-1},$$ where $C_1=G$, $C_{k+1}=G+E_1+\cdots E_k$ for $k=1,\dots,r$, and $v_k=1/(1+\operatorname{tr}(C_{k}^{-1}E_k))$.
As an illustration, the author calculates the inverse of $I+H$ where $I$ is the identity matrix and $H$ has rank two as: $$ (I+H)^{-1}=I-\frac{1}{a+b}(aH-H^2),\qquad \text{($\star$)}$$ where $a=1+\operatorname{tr}H$ and $2b=(\operatorname{tr}H)^2-\operatorname{tr}(H^2).$
The algebra to obtain this result is not obvious to me since we now have a more compact formula instead of the recursive one. How do we get from the recursive formula to Eq. ($\star$)? Second, is it possible to come up with a similar formula for the inverse of $I+H$ where H has any rank $r>1$?