Prove that $\sum\limits_{k}\sum\limits_{i\le k}\binom{n}{i}\cdot\sum\limits_{j>k}\binom{n}{j}=\frac{n}{2}\binom{2n}{n}$ Question:

show  that
$$\sum_{k=0}^{n-1}\left(\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{k}\right)\left(\binom{n}{k+1}+\cdots+\binom{n}{n}\right)=\dfrac{n}{2}\binom{2n}{n}$$

My idea: let $$a_{k}=\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{k},b_{k}=\binom{n}{k+1}+\cdots+\binom{n}{n}$$
so use Abel identity
$$\sum_{k=0}^{n-1}a_{k}b_{k}=S_{n-1}b_{n-1}+\sum_{k=1}^{n-2}S_{k}[b_{k}-b_{k+1}]$$
where $$S_{n}=a_{1}+a_{2}+\cdots+a_{n}$$
then I can't  continue,becase I can't how to deal the $S_{k}[b_{k}-b_{k+1}]$
 A: $$
\begin{align}
&\sum_{k=0}^n\left(\sum_{i=0}^k\binom{n}{i}\right)\left(\sum_{j=k+1}^n\binom{n}{j}\right)\tag{1}\\
&=\sum_{k=0}^n\sum_{i=0}^k\sum_{j=k+1}^n\binom{n}{i}\binom{n}{j}\tag{2}\\
&=\sum_{\substack{i,j=0\\i\lt j}}^n\sum_{k=i}^{j-1}\binom{n}{i}\binom{n}{j}\tag{3}\\
&=\sum_{\substack{i,j=0\\i\lt j}}^n(j-i)\binom{n}{i}\binom{n}{j}\tag{4}\\
&=n\sum_{\substack{i,j=0\\i\lt j}}^n\left[\binom{n}{i}\binom{n-1}{j-1}-\binom{n-1}{i-1}\binom{n}{j}\right]\tag{5}\\
&=n\sum_{\substack{i,j=0\\i\lt j}}^n\left[\binom{n-1}{i}\binom{n-1}{j-1}-\binom{n-1}{i-1}\binom{n-1}{j}\right]\tag{6}\\
&=n\sum_{\substack{i,j=0\\i\lt j}}^n\left[\binom{n-1}{i+1}\binom{n-1}{j}-\binom{n-1}{i-1}\binom{n-1}{j}\right]\tag{7}\\
&=n\sum_{j=0}^n\left[\binom{n-1}{j}\binom{n-1}{j}+\binom{n-1}{j-1}\binom{n-1}{j}\right]\tag{8}\\[6pt]
&=n\left[\binom{2n-2}{n-1}+\binom{2n-2}{n-2}\right]\tag{9}\\[6pt]
&=n\binom{2n-1}{n-1}\tag{10}\\[6pt]
&=\frac n2\binom{2n}{n}\tag{11}
\end{align}
$$
Explanation:
$\:\ (2)$: distribute product over sum
$\:\ (3)$: change order of summation
$\:\ (4)$: evaluate sum in $k$
$\:\ (5)$: $k\binom{n\vphantom{1}}{k}=n\binom{n-1}{k-1}$
$\:\ (6)$: $\binom{n\vphantom{1}}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$ (Pascal recursion)
$\:\ (7)$: reindex the sum of the left term
$\:\ (8)$: telescoping sum
$\:\ (9)$: sum each term using Vandermonde's Identity
$(10)$: $\binom{n\vphantom{1}}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$ (Pascal recursion)
$(11)$: $n\binom{2n}{n}=2n\binom{2n-1}{n-1}$ ( similar to $(5)$ )
A: 
A general principle in such contexts is to try to get rid of the limits of the sums and that, to do so, indicator functions are exquisitely tailored.

In the present case, using the convention that ${k\choose i}=0$ for every integers $(k,i)$ except when $k$ and $i$ are two nonnegative integers such that $k\geqslant i$ (and then ${k\choose i}$ is the usual binomial coefficient, of course), one sees that the LHS of the question is $$S_n=\sum_k\sum_{i\leqslant k}{n\choose i}\sum_{j\gt k}{n\choose j}=\sum_{k,i,j}{n\choose i}{n\choose j}\mathbf 1_{i\leqslant k\lt j}.$$
The RHS above is a triple sum without delimiters (a sum on $\mathbb Z^3$, if you like) hence we can start to work on it.
For each $i\lt j$ there are $j-i$ integers $k$ such that $i\leqslant k\lt j$ hence $$S_n=\sum_{i,j}(j-i){n\choose i}{n\choose j}\mathbf 1_{i\lt j}=U_n-V_n,$$ with $$U_n=\sum_{i,j}j{n\choose i}{n\choose j}\mathbf 1_{i\lt j},\qquad V_n=\sum_{i,j}i{n\choose i}{n\choose j}\mathbf 1_{i\lt j}.$$
Using the identity $j{n\choose j}=n{n-1\choose j-1}$ and the change of variable $j\to j-1$ in the first sum and the identity $i{n\choose j}=n{n-1\choose i-1}$ and the change of variable $i\to i-1$ in the second sum, one gets $$U_n=n\sum_{i,j}{n\choose i}{n-1\choose j}\mathbf 1_{i\leqslant j},\qquad V_n=n\sum_{i,j}{n-1\choose i}{n\choose j}\mathbf 1_{i\lt j-1}.$$ The change of variables $k=n-1-i$, $\ell=n-j$ in $V_n$ transforms the condition $i\lt j-1$ into $\ell\lt k$ hence $$ V_n=n\sum_{k,\ell}{n-1\choose n-1-k}{n\choose n-\ell}\mathbf 1_{\ell\lt k}=n\sum_{k,\ell}{n\choose \ell}{n-1\choose k}\mathbf 1_{\ell\lt k}.$$
Renaming $(\ell,k)$ as $(i,j)$ and comparing the RHS with $U_n$, one sees that all the terms in $U_n-V_n$ cancel except those such that $i=j$, thus, $$S_n=n\sum_{i}{n\choose i}{n-1\choose i}=n\sum_{i}{n\choose i}{n-1\choose n-1-i}.$$ The sum in the RHS is the coefficient of $x^{n-1}$ in the polynomial $$(1+x)^n\cdot(1+x)^{n-1}=(1+x)^{2n-1},$$ hence, finally, $$S_n=n\cdot{2n-1\choose n-1}=\frac12n\cdot{2n\choose n}.$$
