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The sum of square-sums of the first n natural numbers is finding the sum of sum of squares upto n terms. This series finds the sum of each number upto n, and adds this sums to a sum variable.

The sum of square-sum of first 4 natural numbers is −

sum = (1^{2}) + (1^{2} + 2^{2} ) + (1^{2} + 2^{2} + 3^{2}) + (1^{2} + 2^{2} + 3^{2} + 4^{2} ) = 1 + 5 + 14 + 30 = 50

There are two methods to find the sum of square-sum of first n natural numbers.

**1) Using the for loop.**

In this method, we will Loop through to every number from 1 to N and find the square sum and then add this square sum to a sum variable. this method requires an iterations for n numbers, so it would be e time consuming for greater numbers.

#include <stdio.h> int main() { int n = 6; int sum = 0; for (int i = 1; i <= n; i++) sum += ((i * (i + 1) * (2 * i + 1)) / 6); printf("The square-sum of first %d natural number is %d",n,sum); return 0; }

The square-sum of first 6 natural number is 196

**2) Using Mathematical formula**−

Based on finding the nth term and and general formula for the sequence a mathematical formula is derived to find the sum. the formula to find some of square sums of first n natural number is sum = n*(n+1)*(n+1)*(n+2)/12

Based on this formula we can make a program to find the sum,

#include <stdio.h> int main() { int n = 6; int sum = (n*(n+1)*(n+1)*(n+2))/12; printf("The square-sum of first %d natural number is %d",n,sum); return 0; }

The square-sum of first 6 natural number is 196

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