Question

已知数列an中,a1=1,当n≥2时,其前n项和为Sn,满足Sn²=an(Sn-1)？

Answer 1

Sn²=an(Sn-1)
Sn²=[sn-s(n-1)]*（sn-1)=Sn²-sn*sn(n-1)-sn+sn(n-1)
sn-sn(n-1)=-sn*sn(n-1) 两边同除以sn*sn(n-1)
1/sn-1/sn(n-1)=1 数列Sn分之1为等差数列
2,球数列an的通项公式
1/sn-1/sn(n-1)=1
1/（sn-1）-1/sn(n-2)=1
：
1/s2-1/s1= 1等式相加得
1/sn-1/s1=n-1
1/sn=n
sn=1/n
an=sn-sn-1=1/n-1/(n-1）=-1/n(n-1)
[或用 Sn²=an(Sn-1) an= Sn²/(Sn-1) =1/n²/(1-n)/n]=-1/n(n-1)]
[ n=1 an=1,当n≥2时an=-1/n(n-1) ],6,已知数列an中,a1=1,当n≥2时,其前n项和为Sn,满足Sn²=an(Sn-1)
1,求证,数列Sn分之1为等差数列
2,球数列an的通项公式