# C++ program to print the sum of N natural numbers

In this program, you will find the sum of n natural numbers in the C++ program. There are two ways you can calculate 1) add all consecutive numbers from 1 to n. 2) use formula n(n+1)/2.

1. Sum of first 10 natural number is 1 + 2 + 3+ 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55.
2. Use formula: n(n+1)/2 = 10 ( 10 + 1 ) / 2 = 55.

## calculate the sum of n using for loop

``````
#include <iostream>
using namespace std;

int main() {
int num, sum = 0;

cout << "Enter the number: ";
cin >> num;

for (int i = 1; i <= num; i++) {
sum += i;
}

cout << "Sum = " << sum;
return 0;
}
``````
output
```Enter the number: 10
Sum = 55```

Explanation:

• In this program, you have taken a number from the user and stored it in a variable `num`.
• Iterate the for loop from 1 to num, inside `for` loop, add the current iteration to a variable `sum`.
• After the end of a loop print the `sum` value.

## Sum of n using while loop in C++ program

``````
#include <iostream>
using namespace std;

int main() {
int num, sum = 0, i = 1;

cout << "Enter the number: ";
cin >> num;

while(i <= num){
sum += i;
i++;
}

cout << "Sum = " << sum;
return 0;
}
``````
output
```Enter the number: 8
Sum = 36```

time complexity: O(n)

## Calculate the sum of n using the formula

``````
#include <iostream>
using namespace std;

int main() {
int num, sum = 0;

cout << "Enter the number: ";
cin >> num;

sum = num * (num + 1) / 2;

cout << "Sum = " << sum;
return 0;
}
``````
output
```Enter the number: 5
Sum = 15```

By using the above formula time complexity will become O(1).