The pattern is growing horizontally and vertically, with the shape of a ladder. Each square is a step of the ladder. Is symmetrical in its diagonal.
For figure 1, we have 1 square.
For figure 2, we have 2+1 = 3 squares.
For figure 3, we have 3+2+1 = 6 squares.
For figure 4, we have 4+3+2+1 = 10 squares.
We can also look at the pattern as joining a "tower" of squares of size "n" for figure "n".
Figure n will have:
![numberofsquares_n=\sum ^n_{i=1^{}}i=(n(n+1))/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/mnwmcr131dhx3hpm83xgys6hahufy2xav0.png)
For figure 10 we have: 10*11/2 = 110/2 = 55 squares.
It will look like a triangle with 10 squares in the base and 10 squares in the left side.
For figure 55 we will have 1540 squares.
![\begin{gathered} n=55 \\ (n(n+1))/(2)=(55\cdot56)/(2)=(3080)/(2)=1540 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hjdvcvlrjzv3617lceu34cfna6x608ypod.png)
To know if 190 correspond to one figure we can calculate:
![\begin{gathered} (n(n+1))/(2)=190 \\ n(n+1)=190\cdot2 \\ n^2+n=380 \\ n^2+n-380=0 \\ \\ n=(-1\pm√(1-4\cdot1\cdot(-380)))/(2\cdot1)=(-1)/(2)\pm(√(1+1520))/(2) \\ \\ n=(-1)/(2)\pm\frac{√(1521){}}{2}=(-1)/(2)\pm(39)/(2) \\ \\ n_1=(-1+39)/(2)=(38)/(2)=19 \\ n_2=(-1-39)/(2)=(-40)/(2)=-20 \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ywdpwkrbblpk2v8mauyfdcro5xsvsme33r.png)
We have a solution for n=19, so in figure 19 we will have a ladder with 190 squares, as it satisfy the pattern we have figured.