finding $\mathop{\sum\sum}_{0 \leq i < j\leq n}(i+j)\binom{n}{i}\binom{n}{j}$ finding $\displaystyle \mathop{\sum\sum}_{0 \leq i < j \leq n}(i+j)\binom{n}{i}\binom{n}{j}$
expanding sum $\displaystyle (0+1)\binom{n}{0}\binom{n}{1}+(0+2)\binom{n}{0}\binom{n}{2}+\cdots \cdots +(0+n)\binom{n}{0}\binom{n}{n}+(1+2)\binom{n}{1}\binom{n}{2}+(1+3)\binom{n}{1}\binom{n}{3}+\cdots+(1+n)\binom{n}{1}\binom{n}{n}+\cdots +(n-1+n)\binom{n}{n-1}\binom{n}{n}$
wan,t be able to go further , some help me
 A: You may use a diagonal argument. Observe that 
$$\sum_{0\leq i<j\leq n}(i+j)\binom{n}{i}\binom{n}{j}=\sum_{0\leq j<i\leq n}(i+j)\binom{n}{i}\binom{n}{j},$$
therefore
$$\begin{aligned}\sum_{0\leq i<j\leq n}(i+j)\binom{n}{i}\binom{n}{j}&=\frac{1}{2}\sum_{0\leq i，j\leq n，i\neq j}(i+j)\binom{n}{i}\binom{n}{j}\\
&=\frac{1}{2}\sum_{0\leq i,j\leq n}(i+j)\binom{n}{i}\binom{n}{j}-\sum_{i=0}^n i\binom{n}{i}^2.\end{aligned}$$
We have
$$\begin{aligned}\sum_{0\leq i,j\leq n}(i+j)\binom{n}{i}\binom{n}{j}&=\sum_{0\leq i,j\leq n}i\binom{n}{i}\binom{n}{j}+\sum_{0\leq i,j\leq n}j\binom{n}{i}\binom{n}{j}\\
&=2\sum_{i=0}^n\sum_{j=0}^ni\binom{n}{i}\binom{n}{j}\\
&=2\sum_{i=0}^ni\binom{n}{i}\sum_{j=0}^n\binom{n}{j}\\
&=2\sum_{i=0}^ni\binom{n}{i}(2^n)\\
&=2(n2^{n-1})(2^n)\\
&=n2^{2n}.
\end{aligned}$$
On the other hand
$$\sum_{i=0}^n i\binom{n}{i}^2=n\binom{2n-1}{n-1}.$$
For the proof of this identity you can see here.
Therefore
$$\sum_{0\leq i<j\leq n}(i+j)\binom{n}{i}\binom{n}{j}=n2^{2n-1}-n\binom{2n-1}{n-1}.$$
