The pattern is adding 2 stars on top of the previous figure, so figure number 4 would look like this:
* *
* *
* *
* *
Since the pattern is adding 2 stars every time, the explicit formula (for the number of stars) would be
![f(n)=2n](https://tex.z-dn.net/?f=f%28n%29%3D2n)
Since when n=1, i.e. figure 1, we have 2*1=2 stars
when n=2, i.e. figure 2, we have 2*2=4 stars
when n=3, i.e. figure 3, we have 2*3=6 stars
and so on.
The recursive formula lays on the fact that we add 2 stars to the previous figure, which means that the fomula is
![f_n = f_{n-1}+2](https://tex.z-dn.net/?f=f_n%20%3D%20f_%7Bn-1%7D%2B2)
In fact, what we just wrote is simply "the number of stars at a given iteration is the number of stars at the previous iteration, plus 2.