$$\sum_{k=1}^{\infty}x^k=\frac{x}{1-x}\rightarrow\frac{d}{dx}\sum_{k=1}^{\infty}x^k=\sum_{k=1}^{\infty}kx^{k-1}=\frac{1}{(1-x)^2}\rightarrow \sum_{k=1}^{\infty}kx^{k}=\frac{x}{(1-x)^2}$$
Now let $x=\frac{1}{2}$, hence we have $\sum_{k=1}^{\infty}k({\frac{1}{2}})^{k}=\frac{0.5}{(1-0.5)^2}=2$
Now if we take the derivative, we will have
$$\frac{d}{dx}\sum_{k=1}^{\infty}kx^{k}=\sum_{k=1}^{\infty}k^2x^{k-1}=\frac{d}{dx}\left(\frac{x}{(1-x)^2}\right)=\frac{1+x}{(1-x)^3}\rightarrow \sum_{k=1}^{\infty}k^2x^{k}=\frac{x(1+x)}{(1-x)^3}$$
Now if you let $x=\frac{1}{2}$, then you have:
$$\sum_{k=1}^{\infty}k^2\left(\frac{1}{2}\right)^{k}=\frac{0.5(1+0.5)}{(1-0.5)^3}=\frac{1.5}{0.25}=6$$