Question

已知数列{an}的前n项和为Sn,(an-sn-1)^2=sn*sn-1,（n>=2),且a1=1，an>0

已知数列{an}的前n项和为Sn,(an-sn-1)^2=sn*sn-1,（n>=2),且a1=1，an>0
(1)求a2的值，并证明｛sn｝是等比数列
（2）设bn=（-1）log2sn，Tn=b1+b2+...+bn,求Tn
刚刚打错了 改为设bn=（-1）^nlog2sn

Answer 1

(an-S(n-1))^2=Sn*S(n-1)

(Sn - 2S(n-1))^2=Sn*S(n-1)
(Sn)^2-5Sn.S(n-1) + 4(S(n-1))^2 =0
[Sn -S(n-1)] .[Sn - 4S(n-1) ] =0
Sn =4S(n-1)
Sn = 4^(n-1) . S1
= 4^(n-1)
an = Sn -S(n-1)
= 3.4^(n-2)
ie
an = 1 ; n=1
=3.4^(n-2) ; n=2,3,4,...
a2= 3/4

(2)
bn = -log<2>Sn 是不是这样？
b1=0
T1 =b1=0
for n>=2
bn =-log<2>Sn
=-(log<2>3 + 2n-4 )
Tn = b1+b2+...+bn
=-(n-1)log<2>3 + -(n-3)(n-1)