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:
Answer:
age = 10
name = Cynthia
make an f string so the variables appear in the string. There are other ways to do this but I prefer f strings instead of using string concatenation. It causes problems adding 2 strings together most of the time.
print(f'Hi, {name} ! How are you? I know you are {age} old.')
Business and administration
They're (Almost) the same thing, Silly!
It would be the last statement because 3 is greater than 0 but it does not equal 1 or 2 like the first two statements.
