#include<stdio.h>

int main(){
int n,sum=0,i;
scanf("%d",&n);-

if(n>0){
  sum=(n*(n+1))/2;
}
else if(n<0){
        sum=(n*(-1*n+1))/2;
}
printf("%d",sum);


}

Maybe for N < 0 the formula is
SUM = ((2-N)*(N+1))/2

Edited by author 06.11.2017 19:47

The first term is N
The last term is 1
So the average is (1+N)/2

There are (-N) terms below zero
And there are two more terms
They are 0 and one
So there are (2-N) total terms

So the sum is average by total number of terms

Case "N == 0" isn't processed

thank you for your hint!

Edited by author 20.05.2020 08:52

Edited by author 20.05.2020 08:52

But that should not make a difference