Explanation

(d) The $x$ be the total resistance of infinite network of resistance connected to points $A$ and $B$. Therefore the addition of one step of resistances in the infinite network of resistances will not change the total resistance$\ x$ of the network. Therefore equivalent circuit will be as shown in figure. Then total resistances between $A$’ and $B$’ is $x$ given by
$x = 1 + \frac{1 + x}{1 + x} = \frac{1 + 2x}{1 + x}$
$or\ \ x + x^{2} = 1 + 2x$
or $x^{2} - x - 1 = 0\ or\ x = \frac{1 \pm \sqrt{5}}{2} = \frac{1 + \sqrt{5}}{2}$
Since negative value of $R$ is not possible