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

Let us design the algorithm and write the program to find the sum of the following series:

1

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

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:

Read: n

sum =0

for i=1 to n in steps of 1 do

sum = sum + i*i

[End of for i]

Write : sum

Exit.

The complete program to find the sum of the given series is shown below:

Example : Program to add the series 1

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);

}

Non-executable statement

Input

Enter the number of terms

5

sum=0

i= 1 2 3 4 5

sum =0+1

Output

Sum of squares of numbers =55

1

^{2}+2^{2}+3^{2}+4^{2}+....................................................N^{2}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 , }...............N^{2}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 1

^{2}+2^{2}+3^{2}+4^{2}+....................................................N^{2}### 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+1

^{2}+2^{2}+3^{2}+4^{2}+5^{2}Output

Sum of squares of numbers =55

Program to find sum of squares of first N natural numbers in C
Reviewed by Jacob Lefore
on
December 12, 2014
Rating:

## No comments