数列an.a1=1,a1+2a2+3a3+.+nan=(n+1)/2*an+1

a1+2a2+3a3+。+nan=(n+1)/2*an+1a1+2a2+3a3+。+(n-1)an-1=n/2*an相减 nan=(n+1)/2*an+1-n/2*an得an+1/an=3n/(n+1)即an/an-1=3(n-1)/n(n大于等于2)所以an=a1*(a2/a1)*(a3/a2)*。。。*(an/an-1)=[3^(n-1)]/n(n=1时也满足)