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:
Pseudocode
////////////////////////////////////////////////////////////////////////////////////////////////////////////
Integer netElevation(list of elements of type elevation - type and number)
<em>function open</em>
Define running total = 0
for each element from list
<em>loop open</em>
elevation type = element[i].type
if (elevation type == Up)
running total = running total + element[i].number
else
running total = running total - element[i].number
<em>loop close</em>
return running total
<em>function close</em>
True if it is biology or biotech 3/4 but if it anotomy it is false
Of course it is. No one's going to come and track you down for downloading an image from the internet.
In many cases, images may be under some sort of ownership license like copyright, preventing you from modifying and distributing the image as your own. But again, so many people do it anyway, because no one is going to come knocking on your door because you downloaded an image from the internet.
A free trial is a one-time thing. Try using another email address or something.
