a1+2a2+3a3+...+nan=n(n+1)*(n+2),则：a1+2a2+3a3+...+（n-1）×an-1=n(n-1)*(n+1),两式相减：nan=n(n+1)*(n+2)-n(n-1)*(n+1),得an=3n+3所以：a1+a2+a3+...+an=3*(1+2+3+...+n)+3n=3*n(n+1)/2+3n整理得前n项和为:a1+...