1*2+2*3+...+n(n-1)只有n-1项，an=n(n+1)=n^2+n 故前n项和为：Sn=(1^2+1)+(2^2+2)+(3^2+3)+……+(n^2+n) =(1^2+2^2+3^2+……+n^2)+(1+2+3+……+n) =(n(n+1)(2n+1)/6)+(n(n+1)/2) =(n(n+1)(2n+1)/6)+(3n(n+1)/6) =n(n...
