Question

设数列{an},Sn,a1=1,Sn=an+1-1,设bn=2^n/(an+1)(an+1+1),Tn=b1+b2+…+

Answer 1

Sn=a(n+1)-1

S(n-1)=an-1

an=Sn-S(n-1)=a(n+1)-an

a(n+1)=2an

故{an}是公比为2的等比数列

an=a1*2^(n-1)=2^(n-1)

a(n+1)=2^n

bn=2^n/[a(n+1)][a(n+1)+1]=1/(2^n+1)

因bn