An unusal feature of a usual double sum of product of two binomial coefficiens We know that 
$$\sum_{0 \le i \le j \le n}  A_i~ A_j= \frac{\left(\sum_{k=0}^{n} A_k\right) ^2+\sum_{k=0}^{n} A_k^2}{2}~~(1)$$
One can find $$ S= \sum_{0 \le i \le j < \infty} ~ 2^{-i-j}$$
using (1), we get $S=8/3$. We can also find $S$ as below:
$$S=\sum_{j=0}^{\infty} 2^{-j} \sum_{i=0}^{j} 2^{-i}= \sum_{j=0}^{\infty} 2^{-j} \frac{(2)^{-j-1}-1}{2^{-1}-1} =2\sum_{j=0}^{\infty}[2^{-j}-(2)^{-2j-1}]=8/3.$$
Using (1). we usually get $$T=\sum_{0 \le i \le j \le n}  {n \choose i} {n \choose j} = \frac{4^n+{2n \choose n}}{2}.$$
But we cannon get $T$ as $$ T =\sum_{j=0}^{n} {n \choose j} \sum_{i=0}^{j} {n \choose i},$$
because the second term is not doable in a closed form. Any help!
 A: We  use  the coefficient of operator $[z^j]$  to denote the coefficient of $z^j$ in a series. This way we can write for instance
\begin{align*}
[z^j](1+z)^n=\binom{n}{j}\tag{1}
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{j=0}^n}&\color{blue}{\binom{n}{j}\sum_{i=0}^j\binom{n}{i}}\\
&=\sum_{j=0}^n\binom{n}{j}\sum_{i=0}^j[z^i](1+z)^n\tag{2}\\
&=[z^0](1+z)^n\sum_{j=0}^n\binom{n}{j}\sum_{i=0}^jz^{-i}\tag{3}\\
&=[z^0](1+z)^n\sum_{j=0}^n\binom{n}{j}\frac{z^{-j-1}-1}{z^{-1}-1}\tag{4}\\
&=[z^0](1+z)^n\sum_{j=0}^n\binom{n}{j}\frac{z^{-j}-z}{1-z}\\
&=[z^0]\frac{(1+z)^n}{1-z}\left(\sum_{j=0}^n\binom{n}{j}z^{-j}-z\sum_{j=0}^n\binom{n}{j}\right)\\
&=[z^0]\frac{(1+z)^n}{1-z}\left(\left(1+\frac{1}{z}\right)^n-2^nz\right)\tag{5}\\
&=[z^n](1+z)^{2n}\sum_{j=0}^\infty z^j\tag{6}\\
&=\sum_{j=0}^n[z^j](1+z)^{2n}\\
&=\sum_{j=0}^n\binom{2n}{j}\\
&=\frac{1}{2}\left(\sum_{j=0}^{2n}\binom{2n}{j}+\binom{2n}{n}\right)\tag{7}\\
&\,\,\color{blue}{=\frac{1}{2}\left(4^n+\binom{2n}{n}\right)}
\end{align*}
  and the claim follows.

Comment:


*

*In (2) we use the coefficient of operator according to (1).

*In (3) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

*In (4) we use the finite geometric series formula.

*In (5) we apply the binomial theorem twice and observe the term $2^nz$ does not contribute to $[z^0]$.

*In   (6)  we  factor out $\frac{1}{z^n}$ from $\left(1+\frac{1}{z}\right)^n$ and use  again  the rule as in (3). We also expand the geometric series.

*In (7) we use the symmetry $\binom{2n}{j}=\binom{2n}{2n-j}$ and compensate the factor $\frac{1}{2}$ of the middle term $\binom{2n}{n}$ by additionally adding it.
