1到N的平方和公式

12+22+...+n2=n(n+1)(2n+1)/6

可以用數學歸納法證明

1*2 + 2*3 + 3*4 + ... + n*(n+1)
= (12+1) + (22+2) + (32+3) + ... + (n2+n)
= (12+22+32+...+n2) + (1+2+3+...+n)
= n(n+1)(2n+1)/6 + n(n+1)/2
= [n(n+1)/6] * (2n+1+3)
= n(n+1)(n+2)/3