C Program to Find Sum of series 1²+2²+3²+….+n²

How to write a C Program to Find Sum of series 1²+2²+3²+….+n² using For Loop, and Functions with example. Before we get into the C example, The Mathematical formula for C Program to find Sum of series 1²+2²+3²+….+n² = ( n (n+1) (2n+1)) / 6

C Program to Find Sum of series 1²+2²+3²+….+n²

In this C program, the user asked to enter any positive integer. Then using that value, the compiler will find the sum of series 1² + 2² + 3² + … + n² using the above formula.

/* C Program to Calculate Sum of series 1²+2²+3²+....+n² */
#include <stdio.h>

int main()
{
  int Number, Sum = 0;

  printf("\n Please Enter any positive integer \n");
  scanf(" %d",&Number);

  Sum = (Number * (Number + 1) * (2 * Number + 1 )) / 6;
 
  printf("\n The Sum of Series for %d = %d ",Number, Sum);

}

OUTPUT

ANALYSIS

Within the main() function, We declared 2 integer variables Number and Sum. Below printf statement will ask the user to enter any integer value.

 printf("\n Please Enter any positive integer \n");

scanf(" %d",&Number);

In the next line, We are calculating the Sum of the series 1²+2²+3²+4²+5² using above formula
Sum = (Number * (Number + 1) * (2 * Number + 1 )) / 6;
Sum = (5 * (5 + 1) * (2 * 5 +1)) / 6
Sum = (5 * 6 * 11) / 6
Sum = 330 /6
Sum = 55

The C Program to Find Sum of series 1²+2²+3²+….+n² final output for 5 = 55

C Program to calculate Sum of series 1²+2²+3²+….+n²

If you want to display the series order 1² + 2² + 3² +4² + 5² in the output, then we have to add extra For loop to display as below

/* C Program to Calculate Sum of series 1²+2²+3²+....+n² */
#include <stdio.h>

int main()
{
  int Number, i, Sum = 0;

  printf("\nPlease Enter any positive integer \n");
  scanf("%d",&Number);

  Sum = (Number * (Number + 1) * (2 * Number + 1 )) / 6;
  
  for(i =1; i<=Number;i++)
  {
    if (i != Number)
       printf("%d^2 + ",i);
    
    else
       printf("%d^2 = %d ",i, Sum);
  }
}

OUTPUT

ANALYSIS

The for loop inside the main function will traverse each member and displays the output

 for(i =1; i<=Number; i++)
  {
    if (i != Number)
       printf("%d^2 + ",i);
    else
       printf("%d^2 = %d ",i, Sum);
  }

Sum = (Number * (Number + 1) * (2 * Number + 1 )) / 6;
Sum = (4 * (4 + 1) * (2 * 4 +1))/6
Sum = (4 * 5 * 9) / 6
= 180 /6
Sum = 30

Now, the compiler will enter into for loop

First Iteration
i = 1, so the condition inside the for loop (i <= Number) is TRUE (1 <=4). Next, It will go to if condition (i != Number). It means (1 != 4) – Which is TRUE. So, it will print the output as 1²+

i = 2. Repeat the same until i reaches 4. When i = 4, if condition fails. So, the Else statement is printed.

The final Output will be 1²+2²+3²+4² = 30

C Program to Find Sum of series 1²+2²+3²+….+n² using Functions

In this C program, the user enters any positive integer and then using that value, compiler will find the sum of series 1² + 2² + 3² + … + n² using functions.

/* C Program to Calculate Sum of series 1²+2²+3²+....+n² */
#include <stdio.h>

void Sum_Of_Series(int);

int main()
{
  int Number;

  printf("\n Please Enter any positive integer \n");
  scanf("%d",&Number);

  Sum_Of_Series(Number);
}

void Sum_Of_Series(int Number)
{
  int i, Sum;
  Sum = (Number * (Number + 1) * (2 * Number + 1 )) / 6;
  
  for(i =1;i<=Number;i++)
  {
   if (i != Number)
     printf("%d^2&nbsp; + ",i);
   else
     printf(" %d^2 = %d ", i, Sum); 
  } 
}

OUTPUT

Program to Find Sum of series 1²+2²+3²+….+n² using Recursion

Please refer to Find Sum of series 1²+2²+3²+….+n² using Recursion in Recursion Article to see the code and working principle.