Ok, let's call the equivalent resistance Z (to be consistent with patty's notation). So what does Z equal? Z = R + 1/( 1/R + 1/(some stuff) ) That's just a result of a resistor in series with another resistor that's in parallel to some stuff. Now, take a good look at the "some stuff" on the diagram. Look familiar? In fact, "some stuff" is Z! So just substitute Z in for "some stuff" and then solve for Z. A more mathematical but identical solution, you can view the equivalent resistance as a continued fraction. Then you can solve it like any other continued fraction.
phi*R I never said your solution was wrong. I just didn't understand how you got there. You started babbling on about some fibonacci thing which I honestly have no clue how you got to.
that's why I'm telling you to simplify that continued fraction...step by step you should see it then...the convergents comes out to being 2/1 R, 5/3 R, 13/8 R, 34/21 R, etc those are fibonacci numbers: 1, 2, 3, 5, 8, 13, 21, 34 where do you think the phi for the solution comes from? it's because the fibonacci equation is in terms of phi
Oh, I see now. Instead of looking for a fibonacci pattern, I just solved the continued fraction and got the phi from the resulting quadratic.