在数列{an}中,an=SnS(n-1)(n≥2),且a1=2/9 求证数列{1/Sn)是等差数列,求

问题描述:

在数列{an}中,an=SnS(n-1)(n≥2),且a1=2/9 求证数列{1/Sn)是等差数列,求
求数列{1/Sn}前n项和Tn

1.证:n=1时,1/S1=1/a1=1/(2/9)=9/2n≥2时,1/Sn-1/S(n-1)=[S(n-1)-Sn]/[SnS(n-1)]=-an/[SnS(n-1)]an=SnS(n-1)代入,1/Sn-1/S(n-1)=-an/an=-1,为定值.数列{1/Sn}是以9/2为首项,-1为公差的等差数列.2.1/Sn=9/2+(-1)(n-1...