Sum of resistors when in series $r_{serial} = r_{1}+r_{2}$
Sum of resistors when in parallel, $\dfrac{1}{r_{parallel}}= \dfrac{1}{r1} + \dfrac{1}{r2}$
$\implies r_{parallel} = \dfrac{r_{1}\cdot r_{2}}{r_{1}+r_{2}}$
Every node siblings are in parallel and the sum of each level are in series and all node of the last level are tied so all are in series.
So, total sum of resistor $=$ root $ + $ total of level $2 \ + $ total of level $3 + \ldots + $ total of level $n-1 +$ total of level $n$ (in series)
$\begin{array}{cc}\textbf{ Level}&\textbf{Number of nodes}\\ \hline
1&1 (2^0)\\
2&2 (2^1)\\
3 & 4 (2^2)\\4 & 8 (2^3)\\.\\.\\n-1 & 2^{n-2}\\n & 2^{n-1}\\
\end{array}$
So, total sum of resistor $=$ root $+$ total of level $2 \ +$ total of level $3 + \dots.+$ total of level $n-1 + \ $ total of level $n$ (series)
$\qquad= 1 + \frac{r}{2} + \frac{r}{4} + \frac{r}{8} + \frac{r}{16} + \dots + \frac{r}{2^{n-2}} + (\underbrace{r + r + r + r + \dots +r}_{2^{n-1} \ times}) $ (here $r = 1)$
$\qquad = 1 + \left\{ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^{n-2}}\right\}(\text{decreasing GP}) + {(\underbrace{1+1+1+\ldots+1}_{ 2^{n-1} \ times})} $
$\qquad \approx (2^n) \ \text{approx}$
Option $(A)$ is the correct choice.
