Proving two binomial coefficient identities based on the expansion of $(1 + x)^{2n}$ This is a very interesting combinatorics problem that I came across in an old textbook of mine. So I know its got something to do with permutations and combinations, which yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

Use counting arguments to prove these identities:
$$1. \quad\begin{pmatrix}
n \\
0\\
\end{pmatrix}^2 + \begin{pmatrix} n \\ 1\\ \end{pmatrix}^2 +\begin{pmatrix} n \\ 2\\ \end{pmatrix}^2 + \ldots+\begin{pmatrix} n \\ n\\ \end{pmatrix}^2= \begin{pmatrix} 2n \\ n\\ \end{pmatrix}$$
$$2. \quad\ \begin{pmatrix}
n \\
1\\
\end{pmatrix}^2 + 2\begin{pmatrix} n \\ 2\\ \end{pmatrix}^2 +3\begin{pmatrix} n \\ 3\\ \end{pmatrix}^2 + \ldots+n\begin{pmatrix} n \\ n\\ \end{pmatrix}^2= \frac n2\begin{pmatrix} 2n \\ n\\ \end{pmatrix}$$
Use $(1+x)^{2n}$ to prove Identity $1$.
Use Identity $1$ to prove Identity $2$.


Part 1:
$$\begin{align}
(1+x)^{2n}&=\sum_{m=0}^{2n}\binom {2n}mx^m\tag{1}\\
(1+x)^{2n}&=(1+x)^n(1+x)^n\\
&=\sum_{r=0}^n\binom nr x^r\sum_{s=0}^n\binom ns x^s\\
&=\sum_{r=0}^n\sum_{s=0}^n\binom nr\binom nsx^{r+s}\\
&=\sum_{m=0}^{2n}\sum_{r=0}^m\binom nr\binom n{m-r}x^m\tag{2}\\
[x^m](2)=[x^m](1):\qquad 
\sum_{r=0}^m\binom nr\binom n{m-r}&=\binom {2n}m\\
\text{Put }m=n:\qquad
\sum_{r=0}^n\binom nr\binom n{n-r}&=\binom {2n}n\\
\sum_{r=0}^n\binom nr^2&=\binom {2n}n\color{red}\blacksquare
\end{align}$$
Part 2:
$$\begin{align}
(1)=(2):\qquad 
\sum_{m=0}^{2n}\binom {2n}mx^m&=\sum_{m=0|}^{2n}\sum_{r=0}^m\binom nr\binom n{m-r}x^m\\
\frac d{dx}:\qquad 
\sum_{m=0}^{2n}\binom {2n}m mx^{m-1}
&=\sum_{m=0}^{2n}\sum_{r=0}^m\binom nr\binom n{m-r}mx^{m-1}\\
[x^{m-1}]:\qquad \qquad
m\binom {2n}m&=\sum_{r=0}^m m\binom nr\binom n{m-r}\\
\text{Put }m=n:\qquad 
n\binom {2n}n
&=\sum_{r=0}^n n\binom nr\binom n{n-r}\\
&=\sum_{r=0}^n n\binom nr^2\\
&=\sum_{r=0}^n r\binom nr^2+(n-r)\binom nr^2\\
&=\sum_{r=0}^n r\binom nr^2+\sum_{r=0}^n (n-r)\binom n{n-r}^2\\
&=\sum_{r=0}^n r\binom nr^2+\sum_{r'=0}^n r'\binom n{r'}^2
&&(r'=n-r)\\
&=2\sum_{r=0}^n r\binom nr^2\\
\sum_{r=0}^n r\binom nr^2
&=\frac n2\binom {2n}n\color{red}\blacksquare\end{align}$$
