Tetra-series

[andydude]

You need only factor out B or B^-1 in one of the expressions.
Say C = B^-1 for brevity

Code:

.
   (I + B)^-1 + (I + C)^-1  
= (I + B)^-1 + (CB + C)^-1
= (I + B)^-1 + (C(B + I))^-1
= (I + B)^-1 + (B + I)^-1*C^-1
= (I + B)^-1 + (B + I)^-1*B
= (I + B)^-1 *(I + B)
= I

This completes the proof in my view. Good job.

Andrew Robbins