$\begin{align*}
&\Rightarrow y = 1+2\cdot\left ( \frac{1}{2} \right ) + 3\cdot\left ( \frac{1}{4} \right ) + 4\cdot\left ( \frac{1}{8} \right ) + \dots \\
&\Rightarrow y = {\color{red}{\sum_{j\geq1}^{\infty} j\cdot x^{j-1}}} \qquad \left [ x = \frac{1}{2} \right ] \\
&\Rightarrow y = \sum_{j\geq0}^{\infty} (j+1)\cdot x^{j} \qquad \left [ (j-1) \rightarrow j \right ] \\
&\Rightarrow y = \sum_{j\geq0}^{\infty} j\cdot x^{j} + \sum_{j\geq0}^{\infty} x^{j} \\
&\Rightarrow y = x\cdot {\color{red}{\sum_{j\geq1}^{\infty} j\cdot x^{j-1}}} + \sum_{j\geq0}^{\infty} x^{j} \\
&\Rightarrow y = x\cdot y + \sum_{j\geq0}^{\infty} x^{j} \\
&\Rightarrow y = \left ( 1-x \right )^{-1} \cdot \sum_{j\geq0}^{\infty} x^{j} \\
&\Rightarrow y = \frac{1}{\left ( 1-x \right )^{2}} = 4 \\
\end{align*}$
$\begin{align*}
&\Rightarrow y = 1 \cdot x^0 + 2 \cdot x^1 + 3 \cdot x^2 + 4 \cdot x^3 + \dots + \infty \\
&\Rightarrow \int y \cdot dx = x + x^2 + x^3 + x^4 + \dots + \infty \\
&\Rightarrow \int y \cdot dx = x \cdot \left ( 1 + x^1 + x^2 + x^3 + \dots + \infty \right ) \\
&\Rightarrow \int y \cdot dx = \frac{x}{1-x} \\
&\Rightarrow y = \frac{\mathrm{d} }{\mathrm{d} x}\left [ \frac{x}{1-x} \right ] \\
&\Rightarrow y = \frac{\left ( 1-x \right ) \cdot 1 - x \cdot \left ( -1 \right )}{\left ( 1-x \right )^2} \\
&\Rightarrow y = \left [ \frac{1}{\left ( 1-x \right )^2} \right ]_{\left \{ x = \frac{1}{2} \right \}} = 4 \\
\end{align*}$