Sum of series 1^2 + 3^2 + 5^2 + . . . + (2*n – 1)^2

A series is a sequence of numbers that follow a specific mathematical pattern. The series 1² + 3² + 5² + ... + (2*n-1)² represents the sum of squares of the first n odd numbers.

This series has a special property − it follows the formula: Sum = n²(2n² - 1)/3. However, we can also calculate it using loops by adding each term individually.

Syntax

Sum = n²(2n² - 1)/3
// Or using loop: sum += (2*i - 1) * (2*i - 1)

Method 1: Using Loop

This approach calculates the sum by iterating through each odd number and adding its square −

#include <stdio.h>

int main() {
    int i, n, sum = 0;
    n = 7;
    
    for (i = 1; i <= n; i++) {
        sum = sum + (2 * i - 1) * (2 * i - 1);
    }
    
    printf("The sum of series 1^2 + 3^2 + ... + %d^2 is %d<br>", 2*n-1, sum);
    return 0;
}

The sum of series 1^2 + 3^2 + ... + 13^2 is 819

Method 2: Using Formula

For better efficiency, we can use the mathematical formula directly −

#include <stdio.h>

int main() {
    int n, sum;
    n = 7;
    
    sum = (n * n * (2 * n * n - 1)) / 3;
    
    printf("Using formula: Sum = %d<br>", sum);
    printf("Series: ");
    for (int i = 1; i <= n; i++) {
        printf("%d^2", 2*i-1);
        if (i < n) printf(" + ");
    }
    printf(" = %d<br>", sum);
    
    return 0;
}

Using formula: Sum = 819
Series: 1^2 + 3^2 + 5^2 + 7^2 + 9^2 + 11^2 + 13^2 = 819

Comparison

Conclusion

The sum of squares of first n odd numbers can be calculated using either iterative approach or direct formula. The formula method is more efficient with constant time complexity.