When differentiating $\sum\limits_{n=0}^{\infty} \left(\frac{x}{2}\right)^n$ When differentiating $\sum\limits_{n=0}^{\infty}  \left(\frac{x}{2}\right)^n$, gives $\sum\limits_{n=1}^{\infty}  \left(\frac{nx^{n-1}}{2^n}\right)$. so, $n =0$ , becomes $n = 1$. Then, if we were to differentiate $\sum\limits_{n=1}^{\infty}  \left(\frac{x}{2}\right)^n$, does it become $\sum\limits_{n=2}^{\infty}  \left(\frac{nx^{n-1}}{2^n}\right)$? (from $n= 1$, to $n= 2$?) 
If this is how it works, 
is the above equation false? and how could we solve $\sum\limits_{n=1}^{\infty}  \left(\frac{n^2x^n}{2^n}\right)$?

No, because the reason the sum lower bound goes from 0 to 1 is because the derivative kills the first term in the sum. When we differentiate the sum starting from $n=1$, the first term is not killed by the derivative and hence you shouldn't drop it from the sum.
To deal with the sum 
$$
F(x)=\sum_{n=1}^\infty n^2 x^n 2^{-n},
$$
Do the following:
Define
$$
S(x)=\frac2{2-x}=\sum_{n=0}^\infty\left(\frac x2\right)^n.
$$
We have shown
$$
S'(x)=\sum_{n=1}^\infty n\left(\frac x2\right)^n=\frac2{(2-x)^2}.
$$
Similarly, we can show
$$
S''(x)=\sum_{n=2}^\infty n(n-1)\left(\frac x2\right)^n=\frac4{(2-x)^3}.
$$
Finally, note that
\begin{equation}\begin{aligned}
F(x)&=\frac x2+\left(\frac x2\right)^2\sum_{n=2}^\infty n(n-1)\left(\frac{x}2\right)^{n-2}+\frac x 2\sum_{n=2}^\infty n\left(\frac x2\right)^{n-1}\\
&=\frac x2+\left(\frac x2\right)^2S''(x)+\frac x2S(x)-\frac x2\\
&=\left(\frac x2\right)^2\frac4{(2-x)^3}+\frac x2\frac2{(2-x)^2}=\frac{4x}{(2-x)^2}.
\end{aligned}\end{equation}
