Can $n(n+1)2^{n-2} = \sum_{i=1}^{n} i^2 \binom{n}{i}$ be derived from the binomial theorem? Can this identity be derived from the binomial theorem?
$$n(n+1)2^{n-2} = \sum_{i=1}^{n}  i^2 \binom{n}{i}$$

I tried starting from $2^n = \displaystyle\sum_{i=0}^{n} \binom{n}{i}$ and dividing it by $4$
in order to get $2^{n-2}$.

Start from the binomial theorem in the form 
$$(x+1)^n=\sum_{k=0}^n\binom{n}kx^k$$
and differentiate with respect to $x$:
$$\begin{align*}
n(x+1)^{n-1}&=\sum_{k=0}^n\binom{n}kkx^{k-1}\\\\
&=\sum_{k=1}^n\binom{n}kkx^{k-1}\;.
\end{align*}$$
Differentiate again:
$$\begin{align*}
n(n-1)(x+1)^{n-2}&=\sum_{k=1}^n\binom{n}kk(k-1)x^{k-2}\\\\
&=\sum_{k=1}^n\binom{n}kk^2x^{k-2}-\sum_{k=1}^n\binom{n}kkx^{k-2}\;.
\end{align*}$$
Now let $x=1$ to get
$$\begin{align*}
n(n-1)2^{n-2}&=\sum_{k=1}^n\binom{n}kk^2-\sum_{k=1}^n\binom{n}kk\\\\
&=\sum_{k=1}^n\binom{n}kk^2-\sum_{k=1}^n\binom{n-1}{k-1}n\\\\
&=\sum_{k=1}^n\binom{n}kk^2-n\sum_{k=0}^{n-1}\binom{n-1}k\\\\
&=\sum_{k=1}^n\binom{n}kk^2-n2^{n-1}\\\\
&=\sum_{k=1}^n\binom{n}kk^2-2n2^{n-2}\;,
\end{align*}$$
and solve for $\displaystyle\sum_{k=1}^n\binom{n}kk^2$ to get the desired result.
