Difference between sum of the squares of and square of sum first n natural numbers.

Given a number n, find the difference between the square of the sum and the sum of the squares of the first n natural numbers. This is a classic mathematical problem that can be solved efficiently using direct formulas.

Formulas

The two formulas used are −

• Sum of squares of first n natural numbers: n(n+1)(2n+1) / 6
• Sum of first n natural numbers: n(n+1) / 2, then square it

The difference is: (Square of Sum) − (Sum of Squares).

Worked Example

For n = 3 −

Sum of squares = 1² + 2² + 3² = 1 + 4 + 9 = 14

Sum of numbers = 1 + 2 + 3 = 6
Square of sum  = 6² = 36

Difference = 36 - 14 = 22

Java Implementation

The following program calculates the difference using the formulas ?

public class JavaTester {
    public static int difference(int n) {
        // sum of squares of n natural numbers
        int sumSquareN = (n * (n + 1) * (2 * n + 1)) / 6;

        // sum of n natural numbers
        int sumN = (n * (n + 1)) / 2;

        // square of sum of n natural numbers
        int squareSumN = sumN * sumN;

        // difference
        return Math.abs(sumSquareN - squareSumN);
    }

    public static void main(String args[]) {
        int n = 3;
        System.out.println("Number: " + n);
        System.out.println("Difference: " + difference(n));

        n = 10;
        System.out.println("Number: " + n);
        System.out.println("Difference: " + difference(n));
    }
}

The output of the above code is ?

Number: 3
Difference: 22
Number: 10
Difference: 2640

Conclusion

The difference between the square of the sum and the sum of squares can be computed in O(1) time using direct mathematical formulas, without needing loops. The square of the sum always grows faster than the sum of squares, so the difference increases rapidly with n.