Question

设数列｛an｝中,Sn是其前n项和,若首项a1=1,且满足2Sn^2=an（2Sn-1),（n为下标）,(n∈N*,n≥

Answer 1

利用an=Sn-S(n-1)

即2Sn^2=(Sn-S(n-1))(2Sn - 1)

2Sn^2=2Sn^2-2SnS(n-1) -Sn+S(n-1)

2SnS(n-1)+Sn-S(n-1)=0

Sn=S(n-1)/(2S(n-1)+1)

取倒数

1/Sn=2+1/S(n-1)

所以1/Sn是以1/S1=1/a1=1为首项

公差为2的等差数列

所以

1/Sn=1+2(n-1)=2n-1

Sn=1/(2n-1)

n>=2时

an=Sn-S(n-1)

=1/(2n-1)-1/(2(n-1)-1)

=1/(2n-1)-1/(2n-3)

=[(2n-3)-(2n-1)]/[(2n-1)(2n-3)]

=-2/[(2n-1)(2n-3)]

综上

an=1 ,n=1

-2/[(2n-1)(2n-3)] ,n>=2