Squared sum of n odd numbers - JavaScript

We are required to write a JavaScript function that takes in a Number, say n, and finds the sum of the square of first n odd natural Numbers.

For example, if the input number is 3, we need to find the first 3 odd numbers (1, 3, 5) and calculate the sum of their squares:

1² + 3² + 5² = 1 + 9 + 25 = 35

Understanding the Pattern

The first n odd natural numbers follow the pattern: 1, 3, 5, 7, 9... The formula for the i-th odd number is (2 * i) - 1.

Example

const num = 3;
const squaredSum = num => {
    let sum = 0;
    for(let i = 1; i <= num; i++){
        sum += Math.pow((2 * i) - 1, 2);
    };
    return sum;
};
console.log(squaredSum(num));

35

Alternative Approach Using Mathematical Formula

There's also a direct mathematical formula: the sum of squares of first n odd numbers equals n(2n-1)(2n+1)/3.

const squaredSumFormula = n => {
    return n * (2 * n - 1) * (2 * n + 1) / 3;
};

// Test with different values
console.log(squaredSumFormula(3));  // 35
console.log(squaredSumFormula(5));  // 165
console.log(squaredSumFormula(1));  // 1

35
165
1

Comparison

Conclusion

Both methods work effectively. The loop approach is more intuitive, while the mathematical formula provides O(1) time complexity for better performance with large inputs.