Notice how Pattern 2 is Pattern 1 with 4 balls added in the bottom row.
Pattern 3 is Pattern 2 with 5 more balls.
Pattern 4 is Pattern 3 with 6 more balls.
Generalizing the trend, we expect Pattern [tex]n[/tex] to be identical to Pattern [tex]n-1[/tex] with [tex]n+2[/tex] more balls.
If [tex]b_n[/tex] is the number of balls in the [tex]n[/tex]-th pattern, then we have the recursive relation
[tex]\begin{cases} b_1 = 6 \\ b_n = b_{n-1} + n + 2 & \text{for } n>1 \end{cases}[/tex]
We can solve this recurrence by substitution. Using the definition of [tex]b_n[/tex], we have
[tex]b_{n-1} = b_{n-2} + (n-1) + 2 \\\\ \implies b_n = (b_{n-2} + (n-1) + 2) + n + 2 \\\\ \implies b_n = b_{n-2} + 2\times2 + \bigg(n + (n-1)\bigg)[/tex]
[tex]b_{n-2} = b_{n-3} + (n-2) + 2 \\\\ \implies b_n = (b_{n-3} + (n-2) + 2) + 2\times 2 + \bigg(n + (n-1)\bigg) \\\\ \implies b_n = b_{n-3} + 3\times2 + \bigg(n + (n-1) + (n-2)\bigg)[/tex]
and so on, down to
[tex]b_n = b_1 + (n-1)\times2 + \bigg(n + (n-1) + (n-2) + \cdots + 2\bigg)[/tex]
Recall that
[tex]\displaystyle \sum_{i=1}^n i = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2[/tex]
Then we find
[tex]\displaystyle b_n = 6 + 2(n-1) + \sum_{i=2}^n i[/tex]
[tex]\displaystyle b_n = 2n + 4 + \left(\sum_{i=1}^n i - 1\right)[/tex]
[tex]\displaystyle b_n = 2n + 3 + \frac{n(n+1)}2[/tex]
[tex]\displaystyle \boxed{b_n = \frac{n^2+5n+6}2}[/tex]
