Let the equivalent capacitance of the infinite ladder be C.
Because it is an infinite ladder, the change in the equivalent capacitance will be negligible if we add one more ladder at point AB, as shown in the given figure.
The 2μF and C are in series with each other and they are in parallel to 1μF capacitor,
For finding the equivalent capacitance we can write
Ceq=2×C2+C+1
⇒C=2C+2+C2+C
Now solving
(2+C)×C=3C+2
⇒C2−C−2=0
After solving this quadratic equation
C=−1or2
-1 will be neglected
So, the correct answer is 2μF