Calculate the Sum of Natural Numbers

To compute the sum of natural numbers from 1 to n (entered by the user), loops can be used. You will learn how to use for loop and while loop to solve this problem.

The positive numbers 1, 2, 3... are known as natural numbers. The programs below takes a positive integer (let say n) as an input from the user and calculates the sum up to n.

Example #1: Sum of Natural Numbers Using for Loop

#include <stdio.h>
int main()
{
    int n, i, sum = 0;
    
    printf("Enter a positive integer: ");
    scanf("%d",&n);

    for(i=1; i <= n; ++i)
    {
        sum += i;   // sum = sum+i;
    }

    printf("Sum = %d",sum);

    return 0;
}

Example #2: Sum of Natural Numbers Using while Loop

#include <stdio.h>
int main()
{
    int n, i, sum = 0;
    
    printf("Enter a positive integer: ");
    scanf("%d",&n);

    i = 1;
    while ( i <=n )
    {
        sum += i;
        ++i;
    }

    printf("Sum = %d",sum);

    return 0;
}

Output

Enter a positive integer: 100
Sum = 5050

In both programs, the loop is iterated n number of times. And, in each iteration, the value of i is added to sum and i is incremented by 1.

 

Though both programs are technically correct, it is better to use for loop in this case. It's because the number of iteration is known.

The above programs doesn't work properly if the user enters a negative integer. Here's a little modification of the above program to take input from the user until positive integer is entered.

Example #3: Program to Read Input Until User Enters a Positive Integer

#include <stdio.h>
int main()
{
    int n, i, sum = 0;

    do {
        printf("Enter a positive integer: ");
        scanf("%d",&n);
    }
    while (n <= 0);

    for(i=1; i <= n; ++i)
    {
        sum += i;   // sum = sum+i;
    }

    printf("Sum = %d",sum);

    return 0;
}