CPP

sum of all n integers

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

formula for sum of n numbers

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

sum of n natural numbers

//Created by https://ashif.in

#include <bits/stdc++.h>
using namespace std;
int main()
{
	int n;
	cin>>n; //Take Input
	
  	int sum = n*(n+1)/2;    //Calculate Sum
  	
	cout<<sum;		//Print Sum
  
	return 0;
}

n numbers sum

# Not used any loop:
# Time complexity = O(1)
# Space Complexity = O(1)

# n = 10
n = int(input())
print("N-th only even numbers sum =",(n//2)*((n//2)+1))
print("N-th only odd numbers sum =",(n//2)**2)
print("N-th all numbers sum =",n*(n+1)//2)

#output:
# N-th only even numbers sum = 30
# N-th only odd numbers sum = 25
# N-th all numbers sum = 55

sum of n natural numbers

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int n;
	int sum=0;
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		sum+=i;
	}
		cout<<sum<<" ";
	cout<<endl;
	return 0;
}