a1+2a2+3a3+...+nan=n(n+1)(n+2).①
a1+2a2+3a3+...+(n-1)an-1=(n-1)n(n+1).②
①-②得：nan=n(n+1)(n+2)-(n-1)n(n+1)=n(n+1)(n+2-n+1)=3n(n+1)
an=3(n+1)