The sum of the series Σ (2k² - 4) from k = 1 to n can be found using the following formula:
Σ (2k² - 4) = [2(1²) - 4] + [2(2²) - 4] + [2(3²) - 4] + ... + [2(n²) - 4]
= 2(1² + 2² + 3² + ... + n²) - 4n
The sum of squares of the first n natural numbers can be calculated using the formula:
1² + 2² + 3² + ... + n² = [n(n + 1)(2n + 1)] / 6
Substituting this value in the above equation, we get:
Σ (2k² - 4) = 2[(n(n + 1)(2n + 1)) / 6] - 4n
= (n(n + 1)(2n + 1)) / 3 - 4n
Therefore, the sum of the series Σ (2k² - 4) from k = 1 to n is (n(n + 1)(2n + 1)) / 3 - 4n.