Question

已知各项均为正数的数列{an}的前项和为Sn,且Sn,an,1/2成等差数列。

(1)求a1,a2的值；
(2)求数列{an}的通项公式；
(3) 若bn=4-2n（n∈N+），设Cn=bn/an，求数列{cn}的前n项和Tn

麻烦写下具体步骤，谢谢!

Answer 1

1
Sn+1/2=2an
S1=a1 a1+1/2=2a1,a1=1/2
(a2+1/2)+1/2=2a2
a2=1

2
Sn-1+1/2=2an-1
an=Sn-Sn-1=2(an-an-1)
an/an-1=2
an=a1*2^(n-1)
=(1/2)*2^(n-1)=2^(n-2)

3
cn=4/2^(n-2)-2n/2^(n-2)
=16/2^n-n/2^(n-3)
c1=16/2-1/2^(-2)
Tn=(16/2+16/4+...+16/2^n) - (1/2^(-2)+2/2^(-1)+...+n/2^(n-3))
设tn'=(1/2^(-2)+2/2^(-1)+...+n/2^(n-3))
2tn'=(1/2^(-1)+2+3/2^1+..+n/2^(n-2))
(tn'-n/2^(n-3)) -2tn' =1/2^(-2)+1/2^(-1)+1+1/2+..+1/2^(n-2)
tn'=(1/4)(1-1/2^n)/(1-1/2)-n/2^(n-3)
=1/2-1/2^(n+1)-n/2^(n-3)
Tn=16*(1/2)(1-2^n)/(1-1/2) + 1/2-1/2^(n+1)-n/2^(n-3)
=16-2^(n+4)+1/2-1/2^(n+1)-n/2^(n-3)