The first pattern that might come to mind is this:
$\begin{matrix}\underline{1}\end{matrix}\qquad\begin{matrix}1 & \underline{1} \\ \underline{1} & \underline{2}\end{matrix}\qquad\begin{matrix}1 & 1 & \underline{1}\\ 1 & 2 & \underline{2} \\ \underline{1} & \underline{2} & \underline{3}\end{matrix}\qquad\begin{matrix}1 & 1 & 1 & \underline{1}\\ 1 & 2 & 2 & \underline{2}\\ 1 & 2 & 3 & \underline{3} \\ \underline{1} & \underline{2} & \underline{3} & \underline{4}\end{matrix}$
Each array simply adds a ‘shell’ to the previous one, around the righthand side and the bottom. You can make some progress with this pattern, though it’s not the most helpful one: if you add up the new (underlined) bits in each array, you get $1,4,9,16$, which is recognizable as $1^2,2^2,3^2,4^2$. Now the problem suggests computing the sums of the numbers in each array. If we let $s_n$ be the sum of the numbers in the $n$-th array, we have
$\begin{align*} s_1&=1^2\\ s_2&=1^2+2^2\\ s_3&=1^2+2^2+3^2\\ s_4&=1^2+2^2+3^2+4^2 \end{align*}$
It seems pretty clear that these arrays have something to do with sums of consecutive squares. Unfortunately, it’s not clear what, and it’s not obvious that the added ‘shells’ will continue to be consecutive squares. Is there any other way to break up the arrays that involves sums of consecutive squares?
Yes: think of the arrays as representing three-dimensional arrangements of cubical blocks. We’re looking down on them from above, and each number stands for a stack of that many blocks. Thus, the first array represents a single cubical block. The second represents a square of four blocks with a fifth square stacked atop the one in the lower righthand corner; using $\square$ for an empty space and $\blacksquare$ for a blocks, I could ‘draw’ the arrangement like this, where the lefthand side represents the bottom layer of the arrangement and the righthand side, the second layer.
$\begin{array}{c} \blacksquare\;\blacksquare&\qquad\qquad&\square\;\square\\ \blacksquare\;\blacksquare&\qquad\qquad&\square\;\blacksquare \end{array}$
Notice that this just stacks the first arrangment on top of the lower righthand corner of a $2\times 2$ square of blocks.
A similar breakdown by layers (from bottom to top) of the arrangement of blocks represented by the third array looks like this:
$\begin{array}{c} \blacksquare\;\blacksquare\;\blacksquare&\qquad\qquad&\square\;\square\;\square&\qquad\qquad&\square\;\square\;\square\\ \blacksquare\;\blacksquare\;\blacksquare&\qquad\qquad&\square\;\blacksquare\;\blacksquare&\qquad\qquad&\square\;\square\;\square\\ \blacksquare\;\blacksquare\;\blacksquare&\qquad\qquad&\square\;\blacksquare\;\blacksquare&\qquad\qquad&\square\;\square\;\blacksquare \end{array}$
This just stacks the second arrangement on top of the lower righthand corner of a $3\times 3$ square of blocks. Similarly, it’s not hard to see that the fourth arrangement stacks the third on the lower righthand corner of a $4\times 4$ square of blocks:
$\begin{array}{c} \blacksquare\;\blacksquare\;\blacksquare\;\blacksquare&\qquad\qquad&\square\;\square\;\square\;\square&\qquad\qquad&\square\;\square\;\square\;\square&\qquad\qquad&\square\;\square\;\square\;\square\\ \blacksquare\;\blacksquare\;\blacksquare\;\blacksquare&\qquad\qquad&\square\;\blacksquare\;\blacksquare\;\blacksquare&\qquad\qquad&\square\;\square\;\square\;\square&\qquad\qquad&\square\;\square\;\square\;\square\\ \blacksquare\;\blacksquare\;\blacksquare\;\blacksquare&\qquad\qquad&\square\;\blacksquare\;\blacksquare\;\blacksquare&\qquad\qquad&\square\;\square\;\blacksquare\;\blacksquare&\qquad\qquad&\square\;\square\;\square\;\square\\ \blacksquare\;\blacksquare\;\blacksquare\;\blacksquare&\qquad\qquad&\square\;\blacksquare\;\blacksquare\;\blacksquare&\qquad\qquad&\square\;\square\;\blacksquare\;\blacksquare&\qquad\qquad&\square\;\square\;\square\;\blacksquare \end{array}$
When you think about the arrays in this fashion it really is clear that the $n$-th array is just adding a layer of $n^2$ blocks under the $(n-1)$-st array, so that $s_n=1^2+2^2+3^2+\ldots+n^2$, as we suspected from the first pattern.
It may well be that this is all that you were supposed to notice; it would be difficult, for instance, to conjecture a formula for $s_n$ on the basis of these figures. If you already know that formula, $s_n=\frac{n(n+1)(2n+1)}6\;,\tag{1}$ the exercise seems to me fairly pointless, but you can finish it off by observing that since we now know that $s_n$ is the sum of the first $n$ squares, we know that $s_n$ is given by $(1)$.
If not, there are other patterns that you could see, though none that I consider particularly useful. For instance, you could break each array down into the ‘shells’ on the left and top consisting of identical numbers. I’ll use the sixth array as an example.
$\begin{array}{c} 1&1&1&1&1&1\\ 1&2&2&2&2&2\\ 1&2&3&3&3&3\\ 1&2&3&4&4&4\\ 1&2&3&4&5&5\\ 1&2&3&4&5&6 \end{array}$
There are $11$ $1$’s, $9$ $2$’s, $7$ $3$’s, $5$ $4$’s, $3$ $5$’s, and $1$ $6$:
$\begin{array}{r} k:&1&&2&&3&&4&&5&&6\\ \text{Number of }k\text{’s}:&11&&9&&7&&5&&3&&1\\ \hline \text{Total}:&11&+&18&+&21&+&20&+&15&+&6&=&91 \end{array}$
It’s not hard to see that in general the $n$-th array can be broken down similarly into $2n-1$ $1$’s, $2n-3$ $2$’s, and so on down to $1$ $n$. Thus, if we write down the numbers from $1$ through $n$, and under them the odd numbers from $1$ through $2n-1$ in reverse order, multiply the numbers in each column, and add the products across the bottom, we’ll get $s_n$, the sum of the first $n$ squares. For $n=7$, for instance, we get:
$\begin{array}{r} 1&&2&&3&&4&&5&&6&&7\\ 13&&11&&9&&7&&5&&3&&1\\ \hline 13&+&22&+&27&+&28&+&25&+&18&+&7&=&140 \end{array}$
This pattern isn’t very useful $-$ it doesn’t help us to calculate $s_n$ any more easily than by adding up the successive squares, for instance $-$ but it’s a real pattern in this sequence of arrays, one that can be formulated precisely and whose reality can be proved rigorously.