Question

已知数列an的前n项和为Sn,且a1=1,Sn-S(n-1)=2SnS(n-1)

Answer 1

Sn-S(n-1)=2SnS(n-1)
两边同时除以SnS(n-1)
1/S(n-1)-1/S(n)=2
即1/S(n)-1/S(n-1)=-2
故1/S(n)是公差为-2的等差数列
1/S(n)=1/(S1)-2(n-1)=1-2n+2=3-2n
S(n)=1/(3-2n)

a(n)=S(n)-S(n-1)
=1/(3-2n)-1/(5-2n)
=2/(3-2n)(5-2n).............(n>1)

综上
an=2/(3-2n)(5-2n).............(n>1)
an=1..................................(n=1)

Answer 2

an=sn-s(n-1)
sn-s(n-1)=2sns(n-1)
1/sn-1/s(n-1)=(s(n-1)-sn)/sns(n-1)=-2
所以,数列{1/sn}是等差数列,公差是-2.

首项是1/S1=1/a1=1
所以,1/sn=1-(n-1)*2=3-2n
sn=1/(3-2n)

a(n)=S(n)-S(n-1)
=1/(3-2n)-1/(5-2n)
=2/(3-2n)(5-2n)