The circuit diagram shown consists of a large number of element (each element has two resistors R₁ and R₂). The resistance of the resistors in each subsequent element differs by a factor of K = 1/2 from the resistance of the resistors in the previous elements. The equivalent resistance between A and B shown in figure is :

In the given circuit, the first section contains resistors $R_1, R_2$ and in the subsequent sections, the resistors differ by a factor K = 1/2. Let the effective resistance is R. If we exclude 1st section, the remaining will also be an infinite circuit. But for this remaining part, the first section contains resistors $K{R}_1, K{R}_2$ and in the subsequent sections, the resistors differ by a factor K again. Hence by symmetry, the effective resistance will become KR. 

${\mathrm{R}}_{\mathrm{AB}}=\mathrm{R}$, Beyond CD,

${\mathrm{R}}_{\mathrm{eff}}=\mathrm{KR}$

across CD, ${\mathrm{R}}_{\mathrm{net}}=\frac{{\mathrm{R}}_2\left(\frac{\mathrm{R}}{2}\right)}{{\mathrm{R}}_2+\frac{\mathrm{R}}{2}}=\frac{{\mathrm{R}}_2\mathrm{R}}{2{\mathrm{R}}_2+\mathrm{R}}$

$\therefore {\mathrm{R}}_{\mathrm{AB}}={\mathrm{R}}_1+\frac{{\mathrm{R}}_2\mathrm{R}}{2{\mathrm{R}}_2+\mathrm{R}}$

$\Rightarrow R=\frac{{\mathrm{R}}_1\left(2{\mathrm{R}}_2+\mathrm{R}\right)+{\mathrm{R}}_2\mathrm{R}}{2{\mathrm{R}}_2+\mathrm{R}}$

$\Rightarrow 2R_{2}R+R^2=2R_1{R}_2+R_{1}R+R_{2}R$

$\Rightarrow R^2+R\left(R_2-R_1\right)-2R_1{R}_2=0$

$\mathrm{R}=\frac{\left({\mathrm{R}}_1-{\mathrm{R}}_2\right)\pm \sqrt{{\left({\mathrm{R}}_2-{\mathrm{R}}_1\right)}^2+8{\mathrm{R}}_1{\mathrm{R}}_2}}{2}$

$=\frac{\left({\mathrm{R}}_1-{\mathrm{R}}_2\right)+\sqrt{{\mathrm{R}}_1^2+{\mathrm{R}}_2^2+6{\mathrm{R}}_1{\mathrm{R}}_2}}{2}$