Question about a progression?

Question

Could somebody please prove this statement?


Σ n² where n=0 to N-1 = 1/6*[N(N-1)(2N-1)]

Thanks

Answer 1

Proof by induction:

Suppose N=1, then the statement is true as
0² = 0.

Assume the statement is true for (N-1), i.e.
0² + 1² + 2² +...+ (N-2)² = [(N-1)(N-2)(2N-3)]/6

Consider the sum
0² + 1² + 2² +...+ (N-2)² + (N-1)²
= [0² + 1² + 2² +...+ (N-2)² ] + (N-1)²
= [(N-1)(N-2)(2N-3)]/6 + (N-1)²
= (N-1)[(N-2)(2N-3)/6 + (N-1)]
= (N-1)[2N² - 3N - 4N + 6 +6N - 6]/6
= (N-1)[2N² - N]/6
= N(N-1)(2N-1)/6
Q.E.D.

Answer 2

Just sum up the squares by mathematical induction.