关于等差数列的题
问题描述：
关于等差数列的题
设数列{An}的前n项和为Sn,已知A1=1,且满足3Sn^2=An(3Sn-1)(n≥2)
(1)求证:数列{1/Sn}是等差数列
(2)设bn=Sn/3n+1,求数列{bn）的前n项和n
答
(1)
3(Sn)^2=An(3Sn-1),
n≥2,An=Sn-S(n-1),
所以
3(Sn)^2=[Sn-S(n-1)](3Sn-1)
=3(Sn)^2-Sn-3SnS(n-1)+S(n-1)
3SnS(n-1)+Sn-S(n-1)=0,
SnS(n-1)≠0,两边同除以SnS(n-1)
1/Sn-1/S(n-1)=3,
1/S1=1/a1=1,
所以{1/Sn}是以1为首项,公差为3的等差数列.
(2)
由(1)知
1/Sn=1+3(n-1)=3n-2,
所以Sn=1/(3n-2),
Bn=Sn/(3n+1)
=1/[(3n+1)(3n-2)]
=1/3[1/(3n-2)-1/(3n+1)]
所以
Tn=B1+B2+...+Bn
=1/3[1-1/4+1/4-1/7+...-1/(3n-2)+1/(3n-2)-1/(3n+1)]
=1/3[1-1/(3n+1)]
=n/(3n+1)