设数列{an}的前n项和为Sn,且a1=1,Sn=an+1-1(1)求数列{an}的通项公式(2)设bn=2^n设数列{an}的前n项和为Sn,且a1=1,Sn=an+1-1(1)求数列{an}的通项公式(2)设bn=2^n/(an+1)(an+1+1),Tn=b1+b2+…+bn,求证:1/3小于等于Tn小

Question

设数列{an}的前n项和为Sn,且a1 =1,Sn=an+1-1 (1)求数列{an}的通项公式 (2)设bn=2^n
设数列{an}的前n项和为Sn,且a1 =1,Sn=an+1-1
(1)求数列{an}的通项公式
(2)设bn=2^n/(an+1)(an+1+1),Tn=b1+b2+…+bn,求证:1/3小于等于Tn小于1

Answer 1

S1=a1=a2-1
a2=2
Sn=a(n+1)-1
S(n-1)=an-1
a(n+1)=2*an
a1=1
an=2^(n-1)
2^n/(an+1)(an+1+1)
这里能否表达清晰一点,