$\begin{align*} &\text{S} = 1+2*\frac{1}{2}+3*\frac{1}{2}^{2}+4*\frac{1}{2}^{3}+5*\frac{1}{2}^{4} \dots \infty \\ &\Rightarrow \text{S} =\sum_{k=1}^{\infty}\left [ k.{\bf x}^{k-1} \right ] \\ &\Rightarrow \text{S} = 1.x^{0}+2*x^{2-1}+3*x^{3-1}+4*x^{4-1}+5*x^{5-1}\dots\infty \\ &\Rightarrow \int \text{S}dx = x^{1}+x^{2}+x^{3}+x^{4}+x^{5}\dots\infty \\ &\Rightarrow \int \text{S}dx = \frac{1}{1-x} - 1 \\ &\Rightarrow \text{S} = \frac{\mathrm{d} }{\mathrm{d} x}\left [ \frac{1}{1-x} - 1 \right ] \\ &\Rightarrow \text{S} = \frac{(1-x)*0 - 1.(-1)}{(1-x)^2} = \frac{1}{(1-x)^2} = \left ( 1-x \right )^{-2}\\ &\Rightarrow \text{S}_{x=\frac{1}{2}} = \left ( 1-\frac{1}{2} \right )^{-2} \\ \end{align*}$
Or,
$\begin{align*} &\text{One more observation } \\ &\binom{x}{x-1} ={\large x_{\large C_{x-1}}} = x \text{ always} \\ \\ &\text{We have } \dots \\ &\text{S} = 1+2*\frac{1}{2}+3*\frac{1}{2}^{2}+4*\frac{1}{2}^{3}+5*\frac{1}{2}^{4} \dots \infty \\ &\Rightarrow\text{S} = \binom{1}{0}+\binom{2}{1}*\frac{1}{2}+\binom{3}{2}*\frac{1}{2}^{2}+\binom{4}{3}*\frac{1}{2}^{3}+\binom{5}{4}*\frac{1}{2}^{4} \dots \infty \\ &\Rightarrow \text{S} = \sum_{k=0}^{\infty}\left [ \binom{k+1}{k}\left ( \frac{1}{2} \right )^k \right ] \\ &\Rightarrow \text{S} = \sum_{k=0}^{\infty}\left [ \binom{k+2-1}{k}\left ( \frac{1}{2} \right )^k \right ] \\ &\Rightarrow \text{S} = \frac{1}{\left ( 1-x \right )^n} \qquad \text{ n = 2 and x = 1/2} \\ \end{align*}$
Geometric visualization:
$$\begin{align*} S =\text{Total Area}= 2*2 = 4 \end{align*}$$
