Numbers arranged in square arrays - observe patterns, make a conjecture, and prove or disprove it  Please can anyone tell me what the below question means and how to solve it:

Observe patterns. Make a conjecture. Prove or disprove.
$$\begin{matrix}1\end{matrix}\qquad\begin{matrix}1 & 1 \\ 1 & 2\end{matrix}\qquad\begin{matrix}1 & 1 & 1\\ 1 & 2 & 2 \\ 1 & 2 & 3\end{matrix}\qquad\begin{matrix}1 & 1 & 1 & 1\\ 1 & 2 & 2 & 2\\ 1 & 2 & 3 & 3 \\ 1 & 2 & 3 & 4\end{matrix}$$
Compute the sum of all numbers in each square array.


Assuming the pattern is $A_{ij}=\min(i,j)+1$
Each number $m$ in the $n \times n$ matrix is written $2(n-m)+1$ times. So we have
$$\sum_{m=1}^n m(2n-2m+1)$$
$$(2n+1)\sum_{m=1}^n m - 2\sum_{m=1}^n m^2$$
If you know the formula for sums of values and sums of squares, then you're golden.
$$=(2n+1)\cdot n(n+1)/2-n(n+1)(2n+1)/3$$
$$=\frac {n(n+1)(2n+1)} 6$$
It's a sum of squares in disguise. Look: each matrix can be expanded as
$$\begin{matrix}0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}+\begin{matrix}0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1\end{matrix}+\begin{matrix}0 & 0 & 0 & 0\\ 0 & 1 & 1 & 1\\ 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1\end{matrix}+\begin{matrix}1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1\end{matrix}$$
which should make that a little more obvious. Each matrix of size $n \times n$ has a sum of squares up to $n$.
