Math Function Needed?

Question

I need a function that lets me calculate the sum of all positive intergers less than or equal to N, when N is known of course. I know there is one, but I forgot what it was.

Answer 1

The first answer (from "hello") is correct. Here is why it works:

The average of the numbers from 1 to N is equal to the sum of the largest and smallest numbers, divided by 2: (N + 1) / 2

There are N numbers being added.

The sum is equal to the number of values times the average value: N (N + 1) / 2

Answer 2

The function is f(x) = [capital sigma](i=1 to n) i . It's difficult to write math here.

If you want to evaluate this expression, one way is to note that in (1+n)+ (2+(n-1)) + (3+(n-2))+ ... + ((n-1)+2 + (n+1), each number from 1 to n appears exactly twice. The sum is equal to (n+1) + (n+1) + ... + (n+1) n times, since there is a 1-1 correspondence between {1, ..., n} and the first summand in each (). So the sum is n(n+1), so the sum of the numbers from 1 to n is n(n+1)/2.

Answer 3

N(N+1) / 2

Answer 4

Sn=(1/2)n[2a+(n-1)d]