Summation of an array of numbers that grows quadratically in both dimensions (Note:  This question was inspired by Sum a series of series where each value increments by one, and in particular by the discussion in the comments under the answer at https://math.stackexchange.com/a/829458/124095.)
Lets say we have this 4x4 set of series of numbers:
(1, 3, 6, 10)
(2, 5, 9, 14)
(4, 8, 13, 19)
(7, 12, 18, 25)
Knowing that there's no constant increment between numbers, but instead a constant increment of the increment between numbers (its +1), is it possible to make a formula that calculates the sum of this numbers? The sum for the above set is 156, and the m*n set can grow for m,n>=1 .
 A: Let's write your $4\times 4$ set of numbers as a matrix.
$$\begin{bmatrix}1&3&6&10\\2&5&9&14\\4&8&13&19\\7&12&18&25\end{bmatrix}$$
You want to find the sum of all the entries in this matrix.
More generally, you want to know how to find the sum of the entries in larger matrices that follow similar patterns.  What counts as a "similar pattern"?  In your example, each row grows quadratically, as does each column.  Your initial horizontal increment is 2, and the increment of the horizontal increments is 1; your initial vertical increment is 1, and the increment of the vertical increments is also 1.  It's not completely clear what you have in mind for the general case -- do you want to consider examples in which all four of those parameters can be numbers other than 1? -- but let's see what we can do for semi-general case in which the initial horizontal and vertical increments are arbitrary parameters, and the increment of the increments is 1 in both directions.
Such a matrix has the form
$$\begin{bmatrix}
a&a+d&a+2d+1&\cdots&a+nd+T_{n-1}\\
a+k&a+k+d+1&a+k+2d+3&\cdots&a+k+nd+T_n \\
a+2k+1&a+2k+d+3&a+2k+2d+6&\cdots& a+2k+nd+ T_{n+1} \\
\vdots & \vdots & \vdots & & \vdots\\
a+mk+ T_{m-1} & a+mk+d+T_m&a+mk+2d+T_{m+1} & \cdots & a+mk+nd+T_{m+n-1}\end{bmatrix}$$
Here, $d$ is the increment between the first two elements in the first row, and $k$ is the increment between the first two elements in the first column.  $T_n$ denotes the $n^{th}$ triangular number, given by $T_n = \frac{n(n+1)}{2}$.  The particular matrix in the OP has $a=1$, $m=n=3$, $d=2$, and $k=1$.  If we number the columns $0, 1, \dots n$ and we number the rows $0,1, \dots, m$ then the entry in row $i$, column $j$ is $a+ik+jd+T_{j+k-1}$.
Now, before summing the entries, let's first notice that this matrix can be written as the sum of two slightly simpler matrices, as:
$$\begin{bmatrix}
a&a+d&a+2d&\cdots&a+nd\\
a+k&a+k+d&a+k+2d&\cdots&a+k+nd \\
a+2k&a+2k+d&a+2k+2d&\cdots& a+2k+nd\\
\vdots & \vdots & \vdots & & \vdots\\
a+mk & a+mk+d&a+mk+2d & \cdots & a+mk+nd\end{bmatrix} +
\begin{bmatrix}
0&0&1&\cdots&T_{n-1}\\
0&1&3&\cdots&T_n \\
1&3&6&\cdots& T_{n+1} \\
\vdots & \vdots & \vdots & & \vdots\\
T_{m-1} & T_m&T_{m+1} & \cdots & T_{m+n-1}\end{bmatrix}$$
You can think of the first matrix as a "background" matrix that shows you what the numbers in each row and sequence would have been if they had increased linearly; the second matrix is a "correction" matrix that contains the increases in the increments.
Now, the first matrix is easy to sum, using the method in https://math.stackexchange.com/a/829458/124095.  So let's consider that part of the problem solved.
For the second matrix, each row is a sum of consecutive triangular numbers.  Such a sum can be expressed as the difference between two tetrahedral numbers.  Let $P_n$ denote the $n^{th}$ tetrahedral number; then specifically, we have:


*

*Row $0$ sum $= P_{n-1}$

*Row $1$ sum $= P_{n}$

*Row $2$ sum $= P_{n+1}$

*Row $3$ sum $= P_{n+2} - P_1$

*Row $4$ sum $= P_{n+3} - P_2$

*Row $m$ sum $= P_{n+m-1} - P_{m-2}$


Finally, the problem can be finished by summing the row sums, which amounts to summing consecutive tetrahedral numbers.
