数列题目，高分求速解~

a(1)=S(1)=2a(1)-1
a(1)=1
S(n)=2a(n)-n
S(n+1)=2a(n+1)-(n+1)
2a(n)+1=a(n+1)
2[a(n)+1]=[a(n+1)+1]
数列{a(n)+1}是等比数列
a(n)=2^n-1
b(n)=log2(a(n)+1)=n

Cn=2^bn/(an*a(n+1))
=2^n/[(2^n-1)(2^(n+1)-1)]
C1+C2+……+Cn
<=2/3+∑2^n/[(2^n-1)(2^(n+1)-1)]
<2/3+∑2^n/[(2^(2n+1)-2^(n+2))]
<2/3+∑1/(2^(n+1)-4)
=2/3+(1/2)∑1/(2^(n)-2)
=2/3+1/4+(1/2)∑1/(2^(n)-2) 求和n>=3
＝2/3+1/4+(1/2)∑1/2^(n) 求和n>=2
<2/3+1/4+(1/2)(1/2)
=2/3+1/2
<4/3

1.Sn =2an - n
S(n+1) =2a(n+1) -n-1
所以 a(n+1)=1+2an
所以 a(n+1) +1=2(1+an)
即 [a(n+1) +1]/（1+an)）=2
所以 an+1是等比数列，
2.易知: an +1= 2^n
an=2^n - 1
bn= log2(an+1)=n
3.Cn=2^bn/(anXa(n+1))=2^n/[(2^n-1)(2*2^n-1]
=1/(2^n -1) -1/(2*2^n -1)
设前n项和为S
S=1-1/3+1/3-1/7+1/7-1/14.........+1/(2^n -1) -1/(2*2^n -1)
=1-1/(2*2^n -1)<1<4/3
第三问我不知道标准答案怎么做的，这其实是放缩