Finding $\sum_{0 \le r We know that $$\sum_{0 \le r <s \le n} A_r A_s=\frac{1}{2}\left [(\sum_{t=0}^{n} A_t)^2-\sum_{t=0}^{n} A^2_t\right]$$
Nest using $A_r={n \choose r}x^r$, we can get
$$S(x)=\sum_{0 \le r <s \le n} {n \choose r}{n \choose s}x^{r+s}=\frac{1}{2}\left[(\sum_{t=0}^{n} {n \choose r} x^r)^2=\sum_{t=0}^{n} {n \choose t}^2 x^{2t}\right]=\frac{1}{2}\left[(1+x)^{2n}~-~_2F_1(-n,-n,1,x^2)\right]$$ Differentiating w.r.t. $x$ noting that $$\frac{d}{dz}~_2F_1(a,b,c;z) =\frac{ab}{c}~_2F_1(1+a,1+b,1+c;z),$$ we get
$$S'(x)=\sum_{0 \le r <s \le n} (r+s){n \choose r}{n \choose s}x^{r+s-1}=n(1+x)^{2n-1}-n^2x~_2F_1(1-n,1-n,2,x^2)]$$
Let $x=1$, then $$\sum_{0 \le r <s \le n} (r+s) {n \choose r}{n \choose s}=n2^{2n-1}-\frac{n}{2} {2n \choose n}.~~~~~(*)$$
The question is how else one can get this summation (*)?

$$ \begin{align} \tag{1} S &\ = \sum_{0 \le r <s \le n}~ (r+s){n \choose r}{n \choose s} \\
&\ = \sum_{0 \le r <s \le n}~ (r+s){n \choose n-r}{n \choose n-s} \\
&\ \tag{2} = \sum_{0 \le r <s \le n}~ (n-r+n-s){n \choose n-r}{n \choose n-s} \\
\end{align}$$
Add $(1)$ and $(2)$
$$ 2S= \sum_{0 \le r <s \le n}~ 2n{n \choose r}{n \choose s}$$
$$\begin{align} S  &\ = n\sum_{0 \le r <s \le n}~ {n \choose r}{n \choose s} \\ &\ =n \left(2^{2n-1}-\dfrac12 \binom{2n}{n}\right) \\ \end{align} $$

NOTE: Finding $ \sum_{0 \le r <s \le n}~ {n \choose r}{n \choose s}$
$$ (1+x)^{2n}=\sum_{0 \le r,s \le n}~ {n \choose r}{n \choose s}x^{r+s}$$
Put $x=1$
$$ \begin{align} 2^{2n} &\ =\sum_{0 \le r,s \le n}~ {n \choose r}{n \choose s} \\ &\ = \sum_{0 \le r<s \le n}~ {n \choose r}{n \choose s} + \sum_{0 \le s<r \le n}~ {n \choose r}{n \choose s}+ \sum_{0 \le r=s \le n}~ {n \choose r}{n \choose s} \\ 
&\ = S_1 + S_1 +  \sum_{k=0}^{n} \binom{n}{k}^2 \\
&\ = 2S_1 + \binom{2n}{n} \\
\end{align}$$
$$\therefore S_1 = 2^{2n-1} - \dfrac12 \binom{2n}{n}$$
