已知数列{an}的前n项和为Sn,且an=Sn*S(n-1)(n大于等于2,Sn不等于0),a1=2/9(1)求证:{1/Sn}是等差数列(2)求满足an>a(n-1)的自然数n的集合
问题描述:
已知数列{an}的前n项和为Sn,且an=Sn*S(n-1)(n大于等于2,Sn不等于0),a1=2/9
(1)求证:{1/Sn}是等差数列
(2)求满足an>a(n-1)的自然数n的集合
答
an=Sn-S(n-1)所以Sn-S(n-1)=Sn*S(n-1)[Sn-S(n-1)]/Sn*S(n-1)=11/S(n-1)-1/Sn=11/Sn-1/S(n-1)=-1所以1/Sn是等差数列a1=S11/S1=9/21/Sn-1/S(n-1)=-1d=-11/Sn=9/2-(n-1)=-n+11/2=(-2n+11)/2Sn=2/(11-2n)an>a(n-1)Sn-S(n...