1.
n≥2時,
2an=2[Sn-S(n-1)]=2Sn-2S(n-1)
2Sn-2S(n-1)=SnS(n-1)
等式兩邊同除以2SnS(n-1)
1/S(n-1)- 1/Sn=1/2
1/Sn -1/S(n-1)=-1/2
1/S1=1/a1=1/3,數(shù)列{1/Sn}是以1/3為首項,-1/2為公差的等差數(shù)列,公差=-1/2
2.
1/Sn=(1/3)+(-1/2)(n-1)=(5-3n)/6
Sn=6/(5-3n)
n≥2時,an=Sn-S(n-1)=6/(5-3n)- 6/[5-3(n-1)]=6/(5-3n) -6/(8-3n)
n=1時,a1=6/(5-3)-6/(8-3)=9/5≠3
數(shù)列{an}的通項公式為
an=3n=1
6/(5-3n)-6/(8-3n) n≥2
3.
假設(shè)存在滿足題意的k
k=1時,
a2=6/(5-3×2)-6/(8-3×2)=-9
k≥2時,
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
不等式右邊1/(3k-5)²恒>0,要不等式有解,(3k-2)(3k-8)>0k>8/3,k為正整數(shù),k≥3
(3k-2)(3k-8)0,不等式恒成立,即k≥3時,恒滿足題意.
綜上,得k的最小值為3.