设数列an中,a1=1,sn为数列an的前n项和,且an=2Sn2/2Sn-1,求Sn和an

Question

设数列an中,a1=1,sn为数列an的前n项和,且an=2Sn2/2Sn-1,求Sn和an

Answer 1

an＝2Sn^2/2Sn-1,
又an＝Sn-S(n-1),
所以Sn-S(n-1) ＝2Sn^2/2Sn-1,
(2Sn-1)(Sn-S(n-1)) ＝2Sn^2,
2Sn^2-2 Sn S(n-1)- Sn+ S(n-1) ＝2Sn^2,
-2 Sn S(n-1)- Sn+ S(n-1) ＝0,
- Sn+ S(n-1)= 2 Sn S(n-1),
两边同除以Sn S(n-1)可得：1/ Sn -1/ S(n-1)=2,
所以数列{1/ Sn }是等差数列,首项为1,公差为2,
1/ Sn=1+2(n-1)=2n-1,
Sn=1/(2n-1)
∴n=1时,a1=1,
n≥2时,an= Sn- S(n-1)= 1/(2n-1)- 1/(2n-3)=-2/[(2n-1) (2n-3)].
希望能帮到您,