Question

问一道高一数学题,谢谢!!!!

数列{an}的前n项的和为Sn,当n>=1时,Sn+1是an+1与(Sn+1)+2的等比中项
1.求证:数列{1/Sn}为等差数列
2.当a1=-1时,求an

Answer 1

1.
(Sn+1)^2
=an+1*[(Sn+1)+2]
=(Sn+1-Sn)*[(Sn+1)+2]
=(Sn+1)^2-SnSn+1+2(Sn+1-Sn)

所以
SnSn+1=2(Sn+1-Sn)
(1/Sn+1)-(1/Sn)=-1/2
数列{1/Sn}为等差数列,公差为-1/2

2.
S1=a1=-1
1/Sn=-1+(n-1)*(-1/2)=-(1+n)/2
Sn=-2/(1+n)
an=Sn-Sn-1=[-2/(1+n)]-[-2/n]=2/[(n+1)n] (n>=2)
当n=1时,a1=-1

所以
an=-1 (n=1)
an=2/[(n+1)n] (n>=2)