已知数列{An}，An>0(n=1,2,3,……)其前n项和为Sn ,满足(p-1)Sn=p-An,n属于正整数，p>0。

(p-1)Sn=p-an
(p-1)Sn-1=p-an-1
(p-1)an=an-1-an
pan=an-1
an/an-1=1/p
an=a1*(1/p)^(n-1)
(p-1)a1=p-a1
a1=1
an=(1/p)^(n-1)
an=p^(1-n)
bn=1-log(p)(an)
bn=1-1+n=n
Cn=bn/an=2^(1-n)n
Tn=1*2^0+2*2^(-1)+3*2^(-2)+4*2^(-3)+……+n2^(1-n)
Tn/2=1*2^(-1)+2*2^(-2)+3*2^(-3)+4*2^(-4)+……+n2^(-n)
两式相减
Tn/2=2^0+2^(-1)+2^(-2)+2^(-3)+2^(-4)+……+2^(1-n)-n2^(-n)
Tn/2=2-2^(1-n)-n2^(-n)
Tn=4-2^(2-n)-n2^(1-n)
2^(2-n)+n2^(1-n)=2^(2-n)(1+n/2)
0<2^(2-n)(1+n/2)<4
0<4-2^(2-n)-n2^(1-n)<4
0<Tn<4

[I](p-1)Sn=p-an,(1)
(p-1)Sn-1=p-a(n-1),(2)
(1)-(2),得:(p-1)an=a(n-1)-an
pan=a(n-1)
an/a(n-1)=1/p
a(n-1)/a(n-2)=1/p
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a3/a2=1/p
a2/a1=1/p
连乘,得:an/a1=1/p^(n-1)
a1=1
an=1/p^(n-1)为{an}的通项
即an=p^(1-n)
bn=1-log(p)(an)
bn=1-1+n=n为{bn}的通项
[II]Cn=bn/an=n*2^(1-n)
Tn=1*2^0+2*2^(-1)+3*2^(-2)+4*2^(-3)+……+n*2^(1-