Problem Solving

Four concentric circles share the same center. The smallest circle has a radius of 1 inch. For n greater than 1, the area of nth smallest circle in square inches, A_n, is given by $$A_n=A_{n-1}+\left(2n-1\right)\pi.$$
What is the sum of the areas of the four circles, divided by the sum of their circumferences, in inches?

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

The OA is B.

I don't have clear this PS question,

Can I say that,
$$A_1=\pi,\ r_1=1$$
$$A_2=A_1+\ 3\pi=4\pi,\ r_2=2$$
$$A_3=A_2+\ 5\pi=9\pi,\ r_3=3$$
$$A_4=A_3+\ 7\pi=16\pi,\ r_4=4$$
Then, I stuck without ideas about how to continue with the solution.

I appreciate if any expert explain it for me. Thank you so much.

The sum of the areas of the four circles divided by the sum of their circumferences is:

(Π r1^2 +Π r2^2 + Π r3^2 + Π r4^2 )/ (2Π r1 +2Π r2 +2Π r3 + 2Π r4)

Π (r1^2 + r2^2 + r3^2 + r4^2 )/ 2Π ( r1 + r2 + r3 + r4)

(r1^2 + r2^2 + r3^2 + r4^2 )/ 2 ( r1 + r2 + r3 + r4)

You were correct in calculating the radius, so now you just have to plug them in

(1+4+9+16)/2(1+2+3+4)

30/20= 1 ½