在数列an中,a1=1,当n大于2时,前n项和Sn满足Sn的平方=an(Sn-1/2) 求数列an的通项

问题描述:

在数列an中,a1=1,当n大于2时,前n项和Sn满足Sn的平方=an(Sn-1/2) 求数列an的通项

Sn²=an(Sn-1/2)
an=Sn-Sn-1
Sn²=(Sn-Sn-1)(Sn-1/2)
=Sn²-SnSn-1-1/2*Sn+1/2*Sn-1
SnSn-1=-1/2(Sn-S-1)
2SnSn-1=Sn-1-Sn
两边同时除以SnSn-1
(1/Sn)-(1/Sn-1)=2
因为S1=1
S2=1/3
所以1/Sn是首项为1公差为2的等差数列
1/Sn=2n-1
Sn=1/2n-1
Sn-1=1/2n-3
∴an=Sn-Sn-1=(1/2n-1)-(1/2n-3)(可考虑通分)