Here are four different infinite networks. Each one has sections that repeat an infinite number of times connected the same way. For the last one, only one of the sections is shown to save space in the drawing but the others show three such sections.

The idea is to calculate the input impedance Zin for each infinite network.
The results are quite interesting.

Just one note about the last one with ZRRLC. This is not a transmission line, just a bunch of components stuck together and we use the lumped circuit element model rules not physical transmission line rules.

 

Hi again,

I guess nobody wanted to try this yet so i'll give out one of the solutions as well as the method.

Looking at the resistor only circuit with R1 and R2...
First, we look at the circuit with just the three component sections so it's a finite circuit.
Next, we note that R1+R2 is in parallel with the previous R1, and then that combo is in series with the previous R2, and then that series combination is in parallel with the R1 previous to that, and so on and so forth. We can then easily calculate the input impedance.

So what makes the infinite component section version different from the finite one?
The difference is that with the finite one we always knew what went in parallel to the previous R1 so in the end we were able to do the last series combination and so obtain the input impedance without too much trouble. So the key is that we just have to know what to put in parallel with the 'first' R1 on the far right (if we could that is) and then proceed with the calculation as before. Also of interest however is that once we work our way to the left hand side in the finite network, we end up with the same thing: the value that must go in parallel with the 'previous' R1. Since there are no more R1's once we get to the left side, we must have both the required parallel resistance and the input impedance at the same time, and they must be the same. That allows is to write an equation where Rp is the parallel placed resistance and Ri is the input impedance:
Ri=R2+Rp||R1
or:
Ri=R2+Rp*R1/(Rp+R1)

but we also noticed that the parallel placed resistance must be the same as the input resistance, so we end up with:
Ri=R2+Ri*R1/(Ri+R1)

and solving for Ri we get two possible solutions:
Ri=R2/2-(sqrt(R2)*sqrt(R2+4*R1))/2
Ri=(sqrt(R2)*sqrt(R2+4*R1))/2+R2/2

and note the first solution is not valid because we want only positive values for resistances so we end up with:

Ri=(sqrt(R2)*sqrt(R2+4*R1))/2+R2/2

Now with the problem statement values R1=100 and R2=50, the input resistance is 100 Ohms.

Care to try the other ones now?