设bn=a1+2a2+3a3+…+nan则bn=(1+2+3+…n)*2anb(n-1)=(1+2+3+…n-1)*2a(n-1)两式相减得nan=(1+2+3+…n)*2an-(1+2+3+…n-1)*2a(n-1)=n(n+1)an-(n-1)na(n-1)得(n+1)an-(n-1)a(n-1)＝0即an=(n-1)/(n+1)*a(n-1)则an=2/(n+1)n
