I am trying to compute the following finite sum: $$\sum_{i=1}^n\left(A+\lambda_iB\right)^{-1}$$
In this sum, $A$ and $B$ are are positive-definite matrices (so they are inversible). The $\lambda_i$ are positive numbers so the sums $\left(A+\lambda_iB\right)$ are also positive-definite (and inversible).
Is there a way to factor or rewrite this sum in a way where we can only compute $A^{-1}$ and $B^{-1}$ once instead of having to inverse so many matrices?
I tried to use Woodbury's identity but I couldn't simplify the sum much. I could also rewrite the sum as: $$\sum_{i=1}^n B^{-1}\left(B^{-1}+\lambda_iA^{-1}\right)^{-1}A^{-1}$$ but it doesn't really simplify the problem.
Edit: To simplify the problem we can instead compute the following sum: $$S(\lambda) = \sum_{i=1}^n\left(I+\lambda_iC\right)^{-1}$$ where $I$ is the identity matrix and $C$ is another positive-definite matrix.
Actually as $C$ is semi-definite positive, we can diagonalize it and write $C = PDP^\dagger$, where $D$ is diagonal and $P^{-1}=P^\dagger$. So, using some simple matrix algebra:
\begin{align*} (I+\lambda C)^{-1}&=(I+\lambda PDP^\dagger)^{-1} \\ &=\left[(PP^\dagger)(I+\lambda PDP^\dagger)(PP^\dagger)\right]^{-1} \\ &=\left[P(I+\lambda D)P^\dagger\right]^{-1} \\ &=P(I+\lambda D)^{-1}P^\dagger \\ \end{align*}
In this case, the inversion is trivial as $I$ and $D$ are both diagonal matrices and we can easily write the sum without inverting any matrix.