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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.

Sum of first 10 natural number is 1 + 2 + 3+ 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55.
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).

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