Four concentric circles share the same center. The smallest circle has

Official Solution:

Four concentric circles share the same center. The smallest circle has a radius of 1 inch. For \(n\) greater than 1, the area of the \(n\)th smallest circle in square inches, \(A_n\), is given by \(A_n = A_{n-1} + (2n - 1)\pi\).

What is the sum of the areas of the four circles, divided by the sum of their circumferences, in inches?

A. \(1\)
B. \(1\frac{1}{2}\)
C. \(2\)
D. \(2\frac{1}{2}\)
E. \(3\)

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