Answer:
public static double areaSum(Circle c1, Circle c2){
double c1Radius = c1.getRadius();
double c2Radius = c2.getRadius();
return Math.PI * (Math.pow(c1Radius, 2) + Math.pow(c2Radius, 2));
public static void main(String[] args){
Circle c1 = new Circle(6.0);
Circle c2 = new Circle(8.0);
areaSum(c1,c2);
}
Explanation:
An ordered list.
<ol>
<li> This is the first item.
<li> This is the second item.
</ol>
The recursive function would work like this: the n-th odd number is 2n-1. With each iteration, we return the sum of 2n-1 and the sum of the first n-1 odd numbers. The break case is when we have the sum of the first odd number, which is 1, and we return 1.
int recursiveOddSum(int n) {
if(2n-1==1) return 1;
return (2n-1) + recursiveOddSum(n-1);
}
To prove the correctness of this algorithm by induction, we start from the base case as usual:

by definition of the break case, and 1 is indeed the sum of the first odd number (it is a degenerate sum of only one term).
Now we can assume that
returns indeed the sum of the first n-1 odd numbers, and we have to proof that
returns the sum of the first n odd numbers. By the recursive logic, we have

and by induction,
is the sum of the first n-1 odd numbers, and 2n-1 is the n-th odd number. So,
is the sum of the first n odd numbers, as required:

The answer would be D the summary