First, we figure out the area of the smallest circle. $A_1 = \pi r^2 = \pi 1^2 = \pi$.

Now, we find the area of the second smallest circle $(n = 2)$. $A_2 = A_1 + (2(2) - 1) \pi = \pi + 3 \pi = 4 \pi$. This means that the radius of the second smallest circle is 2 (since the area is $\pi r^2$).

The third smallest circle has area $A_3 = A_2 + (2(3) - 1) \pi = 4 \pi + 5 \pi = 9 \pi$. This means that the radius of this circle is 3.

Finally, the fourth smallest circle (that is, the largest circle) has area $A_4 = A_3 + (2(4) - 1) \pi = 9 \pi + 7 \pi = 16 \pi$. This means that the radius of this circle is 4.

The sum of all the areas is $\pi + 4\pi + 9\pi + 16\pi = 30\pi$.

The sum of all the circumferences is $2\pi$ times the sum of all the radii. The sum of all the radii is $1 + 2 + 3 + 4 = 10$, so the circumferences sum up to $20\pi$.

Thus, the sum of all the areas, divided by the sum of all the circumferences, is $\frac{30\pi}{20\pi} = 1\frac{1}{2}$.


Answer: B