Equivalent resistance in ladder circuit

HINT:

Notice that $$R_{eq}=R+\frac1{\frac1R+\frac1{R_{eq}}}$$

You can divide the circuit in n stages, the first stage on the right will show resistance r_0=2R and then you will have to move to the left. At the $n$th stage you will have unknown resistance $r_n$ and at the $(n+1)$th stage you will have resistance: $$r_{n+1}=R+\dfrac{Rr_n}{R+r_n},$$ being the series of $R$ with the parallel between $R$ and $r_n$. This is a recursive succession which, if converges to a limit $L$, it will show $L=r_n=r_{n+1}$ when $n$ tends to infinity. Substituting you will obtain: $$L=R+\dfrac{RL}{R+L}$$ providing the second grade equation $L^2-RL-R^2=0$. The desired limit $L$ is thus the $R_{eq}$ that you are searching: $$R_{eq}=L=\dfrac{R}{2}(1+\sqrt{5})$$