Question

已知数列{an}a1=1 n>=2 Sn^2=an(Sn-1/2)证明是{1/Sn}等差数列？

Answer 1

Sn^2=an[Sn-1/2]
因为an=Sn-S(n-1)代入上式得到
Sn^2=(Sn-S(n-1))[Sn-1/2]
整理S(n-1)-Sn=2SnS(n-1)
左右同时除以SnS(n-1)得到
1/Sn-1/S(n-1)=2
所以{1/Sn}是等差数列,公差为2,首项为1/S1=1,2,an=Sn-S(n-1)
所以(Sn)^2=[Sn-S(n-1)](Sn-1/2)=(Sn)^2-1/2*Sn-Sn*S(n-1)+1/2*S(n-1)
-1/2*Sn-Sn*S(n-1)+1/2*S(n-1)=0
S(n-1)-Sn=2Sn*S(n-1)
[S(n-1)-Sn]/Sn*S(n-1)=2
S(n-1)/Sn*S(n-1)-Sn/Sn*S(n-1)=2
1/Sn-1/S(n-1)=2,是个常数
所以1/Sn是等差数列,0,已知数列{an}a1=1 n>=2 Sn^2=an(Sn-1/2)证明是{1/Sn}等差数列
麻烦大家