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

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