Instead of jumping straight to a sum rule, look at how much the dot count grows from one tile to the next, using the two tiles we already know for certain: the first tile has $2$ dots and the second has $3$ dots, a growth of $1$ dot.
- Growth from tile 1 to tile 2: $3 - 2 = 1$ dot added.
- Expected growth from tile 2 to tile 3 (missing): In this kind of additive picture sequence, the growth itself often grows by matching the previous tile's value, so the next jump is $+2$ (equal to the earlier count of $2$), giving $3 + 2 = 5$ dots for the missing tile.
- Check forward consistency: If the missing tile has $5$ dots, the next jump to the fourth tile should add the previous tile's count again ($+3$), giving $5+3=8$ dots, and the final tile should add $+5$, giving $8+5=13$ dots. This steadily increasing crowd of dots matches what the last tile in the row visually shows, the busiest, most dot-filled tile of the set.
This confirms the missing third tile must carry $5$ dots. Checking the four answer tiles: the $4$-dot square (option A) is short by one dot, the two $6$-dot tiles (options C and D) have one dot too many, and only the $5$-dot tile (option B), with one dot in the center and two dots on either side, lines up with the required count.
Let's summarize:
- Each tile's dot count equals the sum of the two tiles before it, the same building rule as the Fibonacci sequence.
- Starting from $2$ and $3$, the next value has to be $5$.
So the missing tile is the $5$-dot tile, option (B).