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 $2^{n-2}$.
 A: Start with
$$(1+x)^n = \sum_{i=0}^n \binom{n}{i} x^i$$
Take the derivative of both sides:
$$n (1+x)^{n-1} = \sum_{i=1}^n i \binom{n}{i} x^{i-1}$$
Multiply both sides by $x$:
$$n x (1+x)^{n-1}= \sum_{i=1}^n i \binom{n}{i} x^i$$
Take another derivative:
$$n (1+x)^{n-1} + n (n-1) x (1+x)^{n-2} = \sum_{i=1}^n i^2 \binom{n}{i} x^{i-1}$$
Plug in $x=1$ in the above equation:
$$n 2^{n-1} + n (n-1) 2^{n-2} = n (n+1) 2^{n-2} = \sum_{i=1}^n i^2 \binom{n}{i} $$
A: Hint: taking the first two consecutive derivatives
$$(1+x)^n=\sum_{k=0}^nx^k\binom{n}{k}\\n(1+x)^{n-1}=\sum_{k=1}^nkx^{k-1}\binom{n}{k}\\n(n-1)(1+x)^{n-2}=\sum_{k=2}^nk(k-1)x^{k-2}\binom{n}{k}\ldots$$
A: 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.
A: $$\sum_{k=1}^n {n \choose k}k^2= \sum_{k=1}^n \frac n k {n-1 \choose k-1}k^2 $$
$$= \sum_{k=1}^n {n-1 \choose k-1} nk = n \sum_{k=1}^n {n-1 \choose k-1} k $$
$$=n \sum_{s=0}^{n-1} {n-1 \choose s}(s+1) = n \left[\sum_{s=0}^{n-1} {n-1 \choose s}(s)+ \sum_{s=0}^{n-1} {n-1 \choose s}\right]$$
$$=n \left[(n-1)\sum_{s=1}^{n-1} {n-2 \choose {s-1}}+ \sum_{s=0}^{n-1} {n-1 \choose s} \right]$$
By binomial theorem 
$$=n[(n-1)2^{n-2}+2^{n-1}]$$
$$=n(n+1).2^{n-2}$$
A: $$\sum i^2\dbinom{n}{i} = n\left(\sum_{i=1}^n i \dbinom{n-1}{i-1}\right)$$  

since $\dbinom{n}{r}= \dfrac{n}{r}\dbinom{n-1}{i-1}$

$$= n\left(\sum_{i=1}^n(\color{red}{i-1}\color{blue}{+1})\right)\dbinom{n-1}{i-1}\\= n\left(\sum_{i=1}^n(i-1)\dbinom{n-1}{i-1}+\sum_{i=1}^n\dbinom{n-1}{i-1}\right) \\= n\left(\sum_{i=1}^n\dfrac{(n-1)\cancel{(i-1)}}{\require{cancel}\cancel{(i-1)}} \dbinom{n-2}{i-2}+2^{n-1}\right)= n\left((n-1)\sum_{i=1}^n\dbinom{n-2}{i-2}+ 2^{n-1}\right) = n \left[(n-1)2^{n-2}+ 2^{n-1}\right] \\= \color{maroon}{n(n+1)(2^{n-2})}$$
