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(D) Let the equivalent resistance of the infinite network be $x$. Since the network is infinite, adding one more section to the front will not change

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$(\sqrt{3}+1) R$

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(D) Let the equivalent resistance of the infinite network be $x$. Since the network is infinite, adding one more section to the front will not change the equivalent resistance. Thus, the circuit can be represented as a resistor $2R$ in series with the parallel combination of $R$ and $x$.The equivalent resistance $x$ is given by:$x = 2R + \frac{R \cdot x}{R + x}$Multiplying both sides by $(R + x)$:$x(R + x) = 2R(R + x) + Rx$$x^2 + Rx = 2R^2 + 2Rx + Rx$$x^2 - 2Rx - 2R^2 = 0$Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{2R \pm \sqrt{(-2R)^2 - 4(1)(-2R^2)}}{2(1)}$$x = \frac{2R \pm \sqrt{4R^2 + 8R^2}}{2}$$x = \frac{2R \pm \sqrt{12R^2}}{2}$$x = \frac{2R \pm 2R\sqrt{3}}{2}$$x = R(1 \pm \sqrt{3})$Since resistance cannot be negative, we take the positive root:$x = (1 + \sqrt{3})R$

What will be the equivalent resistance between the terminals $A$ and $B$ of the infinite resistive network shown in the figure?

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