Mean square of natural numbers?

This logic works by finding the square of all natural numbers. Find the square of each by looping from 1 to n and add to the sum variable. Then divide that sum by n.

Program for calculating the sum of squares of natural numbers -

Sample code

Real-time demonstration

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

Output

The average of the square of 2 natural numbers is 2.500000

Use the formula to calculate the mean of the squares of natural numbers.

There are mathematical formulas to make calculations easy. To calculate the sum of squares of natural numbers, the formula is ' n*(n 1)*((2*n) 1)/6' Divide it by the number n to get the formula: ' (n 1)* ((2*n) 1 )/6'.

Program for finding the sum of squares of natural numbers -

Sample code

Live demonstration< /p>

#include <stdio.h>
int main() {
   int n = 2;
   float average = ((n+1)*((2*n)+1)/6);
   printf("The average of the square of %d natural numbers is %f", n,average);
   return 0;
}

Output

The average of the square of 2 natural numbers is 2.500000