已知数列{an}满足a1=1,a(n+1)=2an+1(n∈N)

Question

已知数列{an}满足a1=1,a(n+1)=2an+1(n∈N) (1)求数列{an}的通项公式 (2)若数列{bn}满足4^(b1-1).4^(2b2-1).4^(3b3-1)…4^(nbn-1)=(an+1)^n，求数列{bn}的通项公式 (3)若cn=2^n/(ana(n+1))，求数列{cn}的前n项和Sn

Answer 1

a(n+1)=2an+1
a(n+1)+1=2[an+1]
[a(n+1)+1]/[an+1]=2等比公比为2首项2
an+1=2*2(n-1)=2^n
an=2^n-1
(2)
4^(b1+2b2+3b3+……+nbn-n)
=(2^n)^n=2^n*n
2(b1+2b2+3b3+……+nbn-n)=n^2
Bn=b1+2b2+3b3+……+nbn=1/2×n^2+n
Bn-Bn-1=nbn=n+1/2
bn=1/2n+1
(3)若cn=2^n/(ana(n+1))，求数列{cn}的前n项和Sn
cn=2^n/(ana(n+1))=2^n/(2^n-1)[2^(n+1)-1]=1/(2^n-1)-1/[2^(n+1)-1]
Sn=1-1/[2^(n+1)-1]