while loop tutorial – Part 5. Compute sum of number series

Here, we’ll use while loop to compute the sum of a few number series. (You can see part 4 of the while loop tutorial here.)

Q.1. Write a C program to compute 1 + 2 + 3 + … + n, using while loop. Here, n (>=1) is input given by the user.

Answer.

			
//Program to compute the sum of natural numbers from 1 to n, using while loop. 
//Input - n (n>=1) 
//Output - the desired sum 
//Logic - use while loop to iterate the loop counter i from 1 to n. Start with the running sum
//as 0, and increment it by the loop counter in each iteration. 
#include <stdio.h> 
int main()
{
    int sum = 0, i = 1, n; //initially, the running sum is 0. 
                        //loop starts at 1, hence i is set to 1. 
                        //n is the input given by the user 
                        
    printf("Pls enter a number (>=1): ");  //prompt the user to enter a number 
    scanf("%d", &n); //get the input from the user, and save it in the variable n. 
                        
    while (i <= n) //need to compute sum from 1 to n 
    {
        sum += i; //each time we encounter a number, we add it to our running sum  
        i++; //and we increment the number by 1.
    }
    
    //at this point, the variable 'sum' contains the sum 1 + 2 + 3 + ... + n
    printf("The desired sum is %d \n", sum); 
    
    return 0; 
} 

		

			
sample output 1: 
Pls enter a number (>=1): 5
The desired sum is 15
sample output 2: 
Pls enter a number (>=1): 6
The desired sum is 21

		

Explanation:
– We need to compute the sum 1 + 3 + 5 + … + n
– We start the while loop at i = 1. (And set the initial sum to 0.) i keeps getting incremented by 1 in each iteration of the loop. In each iteration, we add the value of i to the running sum. We exit the loop when i goes beyond n.
When we exit the loop, the running sum contains the value of 1 + 2 + 3 + .. + n.

Q.2. Compute the sum of odd numbers till n using while loop. [That is, compute the sum 1 + 3 + 5 + …. + n (or n-1).]

Answer.

			
//Program to compute the sum of odd numbers from 1 to n, using while loop. 
//Input - n (n>=1) 
//Output - 1 + 3 + 5 + ... + n (if n is odd) 
//         1 + 3 + 5 + ... + (n-1) (if n is even)
//Logic - use while loop to iterate the loop counter i from 1 to n. Start with the running sum
//as 0, and increment it by the loop counter in each iteration. Loop counter is incremented by 
// 2 in each iteration (to get to the next odd number)
#include <stdio.h> 
int main()
{
    int sum = 0, i = 1, n; //initially, our sum is 0 
                        //loop starts at 1, hence i is set to 1. 
                        //n is the input given by the user 
                        
    printf("Pls enter a number (>=1): ");  //prompt the user to enter a number 
    scanf("%d", &n); //get the input from the user, and save it in the variable n. 
                        
    while (i <= n) //need to compute sum from 1 to n 
    {
        sum += i; //each time we encounter a number, we add it to our running sum  
        i += 2; //and we increment the number by 2 (to get to the odd number)
    }
    
    //at this point, the variable 'sum' contains the sum 1 + 3 + ... + (n or n-1)
    printf("The desired sum is %d \n", sum); 
    
    return 0; 
}

		

			
sample output 1: 
Pls enter a number (>=1): 5
The desired sum is 9
sample output 2: 
Pls enter a number (>=1): 6
The desired sum is 9
sample output 3: 
Pls enter a number (>=1): 7
The desired sum is 16
sample output 4: 
Pls enter a number (>=1): 8
The desired sum is 16

		

Explanation –
– This is very similar to the previous program. In the previous program, we computed 1 + 2 + 3 + … + n. Here, we compute 1 + 3 + 5 + … .
– We start the while loop at i = 1 as before. The while loop goes till n as before. The initial sum is 0 as before.
– The difference is that we increment i by 2 in each iteration of the loop.
– In each iteration, we add the value of i to our running sum,
– We exit the loop when i goes beyond n.

Q.3. Compute the sum 2 + 5 + 8 + … + (till n), using while loop. (Here n>=2 is an input from the user.)

Answer.

			
//Program to compute the sum 2 + 5 + 8 + ... + (till n), using while loop. 
//Input - n (n>=2) 
//Output - the desired sum 
//Logic - Start the while loop with the loop counter as 2. The running sum is initially 0. 
// In each iteration, increment the running sum by the loop counter. 
// Increment the loop counter by 3 (to get the next number to add to our sum.)
// Continue the loop till n. 
#include <stdio.h> 
int main()
{
    int sum = 0, i = 2, n; 
    //running sum is initially 0. 
    // loop starts at 2. Hence i is set to 2. 
    // n is the input from the user. 
                        
    printf("Pls enter a number (>=2): "); //request the user to enter a number 
    scanf("%d", &n);  //get the input, save it in n. 
    while (i <= n)  //till when to continue the loop? Run the loop till i reaches n. 
    {
        sum += i;  //each number that we encounter, we add it to our running sum 
        i += 3;  //increment the loop counter by 3 to get the next number to add to our sum 
    }
    
    printf("The desired sum is %d \n", sum); 
    
    return 0; 
} 

		

			
output: 
sample output 1:
Pls enter a number (>=2): 5
The desired sum is 7
sample output 2: 
Pls enter a number (>=2): 6
The desired sum is 7
sample output 3: 
Pls enter a number (>=2): 8
The desired sum is 15
sample output 4: 
Pls enter a number (>=2): 10
The desired sum is 15

		

Explanation –
– This is similar to the previous program. Simply the starting number of the series, and the increment value are different.
– We need to compute 2 + 5 + 8 + … + (till n).
– We start the while loop at i = 2. In each iteration of the loop, i is incremented by 3. We continue the loop till i goes beyond i.
– The running sum is initially 0.
– in each iteration of the loop, we add the value of i to our running sum.
– we exit the loop when i becomes greater than n.

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