等差数列的平方求和

An=n^2 Sn=1^2+2^2+3^3+....+n^2
(n+1)(n+2)(n+3)-n(n+1)(m+2)=3(n^2+3n+2)
n^2=[(n+1)(n+2)(n+3)-n(n+1)(n+2)]/3 -3n-2
Sn=[2*3*4-1*2*3+3*4*5-2*3*4+.....+(n+1)(n+2)(n+3)-n(n+1)(n+2)]/3 -3(1+n)*n/2 -2n
=(n^2+3n+2)(n+3)-6 /3 -(3n^2+3n)/2 -2n
=2*(n^3+6n^2+11n+6-6)-9n^2-9n -12n /6= 2n^3+3n^2+n/6=n(2n+1)(n+1)/6

n(n+1)(2n+1)/6
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