In Homework1 , Problem 1-2 , how do I get the effective resistance for the 3rd circuit (the one which has continuous resistance blocks being attached). I am getting a series and not able to make out general expression for the effective resistance. Curious to know the answer. Thank you

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- KenLJW

If you haven't realize yet you can factor all common multiple of resistance from circuit, find a solution and then multiply by the common resistance to fine answer.
There's a number of way's to find solution, I use the Thevinin equivatent resistance for each node with the input shorted.
Looking back from first node Rth = 1/2
looking back from second node Rth = (1/2 + 1)||1 = 3/5
looking back from third node Rth = (3/5 + 1)||1 = 8/13
going to infinite Rth = 1
result is (1 + 1)||1 = 2/3
therefore equivalent resistance is (2/3) R

- KenLJW

This is not quite right, should be greater than R.
If you assume the above value for to the right of node 1, same network analysis for infinite circuit, than the input ressistance is
1 + 1|(2/3) = 7/5
input resistance is (7/5) R
You should check this with others or by doing it another way, please let me know.

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