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.

 A: The other answers are telling you more-or-less how to solve the question, but 
it seems that you are also unsure about what the question means.  Here is what it means:


*

*figure out the pattern in the squares, so that you can correctly draw the fifth, sixth, etc., square.

*Once you know what the pattern is for the squares, write down the sum of all the numbers in the first square, the second square, etc., and figure out a pattern in those numbers.   Typically, this would mean writing down a formula, depending on $n$, whose value at $n$ is equal to the sum of the numbers in the $n$th square.

*Finally, prove that your formula is correct.
A: 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 $n$.
A: Look at the fourth square. The sum of the numbers in row 3 is equal to the sum of row 4 minus 1. The sum of row 2 is the sum of row 4 minus 3. The sum of row 1 is the sum of row 4 minus 5. Do you see a pattern now?
A: Notice what the difference is between each successive square. The second square is the first square plus a row and column. The third square, again, is the second square plus a row and a column. Those new rows and columns follow a simple pattern. In the second square, we're adding 1, 2, 1. In the third square, we're adding 1, 2, 3, 2, 1. And so on. The intuition should be clear that these new terms we are adding each time sum to $n^2$, with $n$ being the number of rows (or columns) in the square.
