1^2+2^2+3^2+……+n^2=n(n+1)(2n+1)/6
自然数列立方和公式 1^3+2^3+3^3+…+n^3=[n^2*(n+1)^2]/4
a(n)=2n^3+3n^2+n
Sn=a1+a2+...+an
=2*(1^3+2^3+...+n^3)+3*(1^2+2^2+...+n^2)+(1+2+3+...+n)
=2*[n^2*(n+1)^2]/4 + 3* [n(n+1)(2n+1)/6]+n(n+1)/2
=n(n+1)(1+2n+1+n^2+n)/2
=n*(n+1)^2*(n+2)/2