sum of n natural numbers

1# Sum of natural numbers up to num
2
3num = 16
4
5if num < 0:
6   print("Enter a positive number")
7else:
8   sum = 0
9   # use while loop to iterate until zero
10   while(num > 0):
11       sum += num
12       num -= 1
13   print("The sum is", sum)
14

1Sum of n integers 1 + 2 + 3 + ... + n = n * (n + 1) / 2

1#include<bits/stdc++.h>
2using namespace std;
3int main()
4{
5	int n;
6	int sum=0;
7	cin>>n;
8	for(int i=1;i<=n;i++)
9	{
10		sum+=i;
11	}
12		cout<<sum<<" ";
13	cout<<endl;
14	return 0;
15}

1Sum of the First n Natural Numbers. We prove the formula 1+ 2+ ... + n = n(n+1) / 2, for n a natural number

1int addNumbers(int n)
2{
3    int count, sum = 0;
4
5    for(count=1; count <= n; count++)
6    {
7        sum = sum + count;
8    }
9  
10    return sum;
11}