数列『an』的前n项和Sn与第n项an之间的关系满足2×lg二分之（Sn-an+1）=lgSn+lg（1-an）.求an和Sn.

Question

数列『an』的前n项和Sn与第n项an之间的关系满足2×lg【二分之（Sn-an+1）】=lgSn+lg（1-an）.求an和Sn.

Answer 1

2lg[(Sn-an+1)/2]=lg[Sn*(1-an)]
an<1
(Sn-an+1)^2=4Sn*(1-an)
Sn^2+an^2+1-2Snan+2Sn-2an=4Sn-4Snan
(Sn+an-1)^2=0,Sn+an=1
S1=a1,a1=1/2
S2=S1+a2,S2+a2=S1+2a2=1,2a2=1-1/2=1/2
a2=1/4
S3+a3=S2+2a3=1,a3=(1-1/4)/2=3/8
S4+a4=S3+2a4=1,a4=(1-S3)/2=(1-3/8)/2=5/16
an=[2^n+(-1)^(n+1)]/[3*2^n]
Sn=1-[2^n+(-1)^(n+1)]/[3*2^n]