$$\sum_{k=0}^n k^2\cdot{n\choose k} = n\cdot(n+1)\cdot2^{n-2}$$
so, since the question had already provided a hint, I planned on expanding using binomial expansion and then differentiating both sides to get my result.
I am stuck thinking of a suitable polynomial that will start the problem. Is there a way to calculate it rather than think about it intuitively?
Let $F(x)=(1+x)^n=\sum_{k=0}^n {n\choose k}x^k$
Then $$(xF'(x))'=n(n-1)x(1+x)^{n-2}+n(1+x)^{n-1}=\sum_{k=0}^n k^2{n\choose k}x^{k-1}$$
Substituting $x=1$ we get the result
In $ (1+x)^n\,=\,\sum_{r=0}^n \binom{n}{r} \,x^r $. $\\$ Different both side by x $\\$ $ n(1+x)^{(n-1)}= \sum_{r=0}^{n} r.\binom{n}{r} x^{(r-1)} $ $\\$ In above equation multiply by x and again different we will get $\\$ $n((1+x)^{(n-1)}+(n-1)x(x+1)^{(n-2)})=\sum_{r=0}^{n} \ r^2\binom{n}{r} x^{(r-1)} $. $\\$ Put x=1 you get $\\$ $ n(n+1)2^{(n-2)}=\sum_{r=0}^{n} r^2\binom{n}{r} $
You can also calculate the identity just using the binomial formula and
\begin{eqnarray*} \sum_{k=0}^n k^2\cdot{n\choose k} & \stackrel{(\star)}{=} & n\sum_{k=\color{blue}{1}}^n k\cdot{n-1\choose k-1} \\ & = & n\sum_{k=\color{blue}{0}}^{\color{blue}{n-1}} (k+1)\cdot{n-1\choose k} \\ & = & n\sum_{k=\color{blue}{0}}^{\color{blue}{n-1}} k\cdot{n-1\choose k} + \underbrace{n\sum_{k=\color{blue}{0}}^{\color{blue}{n-1}} \cdot{n-1\choose k}}_{=n2^{n-1}}\\ & \stackrel{(\star)}{=} & n(n-1)\underbrace{\sum_{k=\color{blue}{1}}^{\color{blue}{n-1}} \cdot{n-2\choose k-1}}_{=2^{n-2}} + n2^{n-1}\\ & = & n(n-1)2^{n-2} + 2\cdot 2^{n-2} = n(n+1)2^{n-2} \end{eqnarray*}
Write $k^2=k(k-1)+k$
If it were $k^3,k^3=k(k-1)(k-2)+ak(k-1)+bk$ where $a,b$ arbitrary constants and can be found to be $b=1,a=3$ by setting $k=1,2$
Now for $k\ge r,$
$$k(k-1)\cdots(k-(r-1))\binom nk$$
$$=k(k-1)\cdots(k-(r-1))\dfrac{n(n-1)\cdots(n-(r-1))}{k(k-1)\cdots(k-(r-1))}\binom {n-r}{k-r}$$
$$=n(n-1)\cdots(n-(r-1))\binom{n-r}{k-r}$$
Finally use $$(1+1)^m=\sum_{p=0}^m\binom mp$$
