Find a closed expression for the sum of the entries of the Pascal triangle inside the upper n x n rhombus. Find a closed expression for the sum of the entries of 
the Pascal triangle inside the upper $n \times n$ rhombus.
For example, for $n = 3$, you need to sum the entries:
\begin{array}{cccccccc}
&&&&1\\
&&&1&&1\\
&&1&&2&&1\\
&\cdot&&3&&3&&\cdot\\
\cdot&&\cdot&&6&&\cdot&&\cdot
\end{array}
I found this problem was really interesting but I was struggling to solve this problem.
I found that the next number below $6$ is $2(1+3+6)$, double the sum of the numbers of the edge of the rhombus. Also, $6=2(1+2)$. I also found that the $m$th number on the edge of the $n \times n$ rhombus is the sum of the first $m$ numbers on the edge of the $(n-1) \times (n-1)$ rhombus. However, I couldn't figure out how to sum them up.
 A: $\newcommand{\cb}{\color{brown}}$You can get it from the hockey stick (or Christmas stocking) identity, which is identity $(8)$ here when $j=k$ in that identity. You have
$$\begin{align*}
\sum_{k=0}^{n-1}\sum_{m=0}^{k+n-1}\binom{m}k&=\sum_{k=0}^{n-1}\binom{k+n}{k+1}\\\\
&=\sum_{k=0}^{n-1}\binom{k+n}{n-1}\\\\
&=\sum_{k=n}^{2n-1}\binom{k}{n-1}\\\\
&=\binom{2n}n-\binom{n-1}{n-1}\\\\
&=\binom{2n}n-1\;.
\end{align*}$$
For example, when $n=3$ it’s $\binom63-1=19$, as can be seen from the figure in the question.
Geometrically this is quite elegant: just subtract $1$ from the binomial coefficient directly below the bottom point of the rhombus.
Added: It may be easier to see why the sum of the entries in the rhombus is the double summation above if you square up Pascal’s triangle:
$$\begin{array}{c|cc}
m\backslash k&0&1&2&3&4&5&6\\ \hline
0&\cb1&0&0&0&0&0&0\\
1&\cb1&\cb1&0&0&0&0&0\\
2&\cb1&\cb2&\cb1&0&0&0&0\\
3&1&\cb3&\cb3&1&0&0&0\\
4&1&4&\cb6&4&1&0&0\\
5&1&5&10&10&5&1&0\\
6&1&6&15&20&15&6&1
\end{array}$$
I’ve colored the $3\times 3$ rhombus brown.
