求一道数列题已知数列an的首项a13,通项an与前n项和Sn满足2an=Sn*S(n-1),（1)求证1/Sn是等差数列

1。n≥2时,2an=2[Sn-S(n-1)]=2Sn-2S(n-1)2Sn-2S(n-1)=SnS(n-1)等式两边同除以2SnS(n-1)1/S(n-1)- 1/Sn=1/21/Sn -1/S(n-1)=-1/21/S1=1/a1=1/3,数列{1/Sn}是以1/3为首项,-1/2为公差的等差数列,公差=-1/22。1/Sn=(1/3)+(-1/2)(n-1)=(5-3n)/6Sn=6/(5-3n)n≥2时,an=Sn-S(n-1)=6/(5-3n)- 6/[5-3(n-1)]=6/(5-3n) -6/(8-3n)n=1时,a1=6/(5-3)-6/(8-3)=9/5≠3数列{an}的通项公式为an=3 n=1 6/(5-3n)-6/(8-3n) n≥23。假设存在满足题意的kk=1时,a2=6/(5-3×2)-6/(8-3×2)=-9k≥2时,ak>a(k+1)6/(5-3k) -6/(8-3k)>6/[5-3(k+1)]-6/[8-3(k+1)]1/(3k-8)+1/(3k-2)>2/(3k-5)1/[(3k-2)(3k-8)]>1/(3k-5)²不等式右边1/(3k-5)²恒>0,要不等式有解,(3k-2)(3k-8)>0 k>8/3,k为正整数,k≥3(3k-2)(3k-8)0,不等式恒成立,即k≥3时,恒满足题意。综上,得k的最小值为3。