Question

已知数列{an}满足a1=1,a(n 1)=2an 1(n∈N*). 若数列{bn}满足4∧b1-1 4∧b2-1……4∧bn-

已知数列｛an｝满足a1=1,a(n+1)=2an+1(n∈N*)。
若数列｛bn｝满足4∧b1-1 4∧b2-1……4∧bn-1=(an+1)∧bn(n∈N*),b5=6求数列｛bn｝是等差数
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Answer 1

a(n+1)=2an+1
a(n+1)+1 = 2(an+1)
[a(n+1)+1]/(an+1) =2
(an+1)/(a1+1)=2^(n-1)
an+1=2^n
an =-1+2^n

4^(b1-1). 4^(b2-1)...4^(bn-1)=(an+1)^bn
4^(b1+b2+..+bn -n) = 2^(nbn)
2^[2(b1+b2+..+bn -n)] = 2^(nbn)
2(b1+b2+..+bn -n) = nbn

let
Sn=b1+b2+...+bn
2(Sn-n) = nbn
n=5
2(S5-5) =5b5
S5= 20

2(Sn-n) = nbn
=n(Sn-S(n-1))
(n-2)Sn =nS(n-1)-2n
Sn/[n(n-1)] - S(n-1)/[(n-1)(n-2)] = -2/[(n-1)(n-2)]
= -2[1/(n-2) - 1/(n-1)]
Sn/[n(n-1)] - S5/[5(4)] = -2[1/4 - 1/(n-1)]
Sn/[n(n-1)] - 1 = -2[1/4-1/(n-1) ]
Sn/[n(n-1)] = 1/2+ 2/(n-1) = (n+3)/[2(n-1)]
Sn = n(n+3)/2
bn = Sn -S(n-1)
= (1/2)[ n(n+3) - (n-1)(n+2)]
= (1/2) [ 2n+2]
=n+1
=>{bn} 是等差数列