Program to find sum of squares of first N natural numbers in C

Jacob
Turbo C - December 12, 2014

Let us design the algorithm and write the program to find the sum of the following series:
1²+2²+3²+4²+....................................................N²
We know the program segment to add all the numbers from 1 to N. The program segment is:
sum=0;
for(i=1; i<=n; i++)
{

sum = sum+i;

}
But, to add 1^{2 ,}2^{2 ,}3^{2 ,}4^{2 ,}...............N² it is necessary to replace the statement:
sum = sum+i;

by the statement:

sum = sum+i*i;

The algorithm and flowchart to find the sum of squares of all natural numbers are as follows:

Algorithm : SERIESSUM

Step 1: [Input the number of terms]
Read: n
Step 2: [Initialization]
sum =0
Step 3: [Find the sum of all terms]
for i=1 to n in steps of 1 do
sum = sum + i*i

[End of for i]

Step 4: [Output the sum]
Write : sum
Step 5: [Finished]
Exit.

The complete program to find the sum of the given series is shown below:
Example : Program to add the series 1²+2²+3²+4²+....................................................N²

PROGRAM

#include<stdio.h>
main( )
{
int n, i, sum;
printf("Enter the number of terms \n");
scanf("%d",&n);
sum =0;
for(i=1; i<=n; i++)
{
sum = sum+(i*i);
}
printf("Sum of squares of numbers = %d",sum);
}

TRACING

Execution starts from here:
Non-executable statement
Input
Enter the number of terms
5
Computations
sum=0
i= 1 2 3 4 5
sum =0+1²+2²+3²+4²+5²
Output
Sum of squares of numbers =55